How do you solve using the quadratic form? 2x^2+3x-3=-3x-4

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How do you solve using the quadratic form? 2x^2+3x-3=-3x-4

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Answer 1 · 2018-05-08T13:49:04

Do you mean the quadratic formula?
$x = - 0.1771243445 or x = - 2.822875656$ $x=-0.1771243445 or x=-2.822875656$

Explanation:

$2 x^{2} + 3 x - 3 = - 3 x - 4$ $2x^2+3x-3=-3x-4$ collect like terms on the LHS

$2 x^{2} + 6 x + 1 = 0$ $2x^2+6x+1=0$

$a = 2, b = 6 and c = 1$ $a=2, b=6 and c=1$
$x = \frac{- 6 \pm \sqrt{6^{2} - [4 \times 2 \times 1]}}{2 \times 2}$ $x=[-6\pmsqrt(6^2-[4xx2xx1])]/[2xx2]$

$x = \frac{- 6 \pm \sqrt{36 - 8}}{4}$ $x=[-6\pmsqrt(36-8)]/[4]$

$x = \frac{- 6 \pm \sqrt{28}}{4}$ $x=[-6\pmsqrt28]/[4]$

$x = \frac{- 6 \pm 2 \sqrt{7}}{4}$ $x=[-6\pm2sqrt7]/[4]$

$x = \frac{- 3 + \sqrt{7}}{2}$ $x=(-3+sqrt7)/2$ or $x = \frac{- 3 - \sqrt{7}}{2}$ $x=(-3-sqrt7)/2$

$x = - 0.1771243445 or x = - 2.822875656$ $x=-0.1771243445 or x=-2.822875656$

Answer 2 · 2018-05-08T13:53:57

$2 x^{2} + 3 x - 3 = - 3 x - 4$ $2 x^2+3 x-3=-3 x-4$

Explanation:

$a x^{2} + b x + c = 0$ $a x^2+b x+c=0$
$x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 \cdot a \cdot c}}{2 \cdot a}$ $x_(1,2)=(-b +- \sqrt(b^2-4*a*c))/(2*a)$

Write your equation in the form: $a x^{2} + b x + c = 0$ $a x^2+b x+c=0$
$2 x^{2} + 3 x - 3 = - 3 x - 4$ $2 x^2+3 x-3=-3 x-4$ | $+ 3 x$ $+3x$
$2 x^{2} + 3 x - 3 + 3 x = - 3 x - 4 + 3 x$ $2 x^2+3 x-3 +3x=-3 x-4 +3x$ | $+ 4$ $+4$
$2 x^{2} + 6 x - 3 + 4 = - 4 + 4$ $2 x^2+6 x-3+4=-4 +4$
$2 x^{2} + 6 x + 1 = 0$ $2 x^2+6 x+1=0$

$a = 2$ $a=2$
$b = 6$ $b=6$
$c = 1$ $c=1$

$x_{1, 2} = \frac{- 6 \pm \sqrt{6^{2} - 4 \cdot 2 \cdot 1}}{2 \cdot 2} =$ $x_(1,2)=(-6 +- \sqrt(6^2-4*2*1))/(2*2)=$
$= \frac{- 6 \pm \sqrt{36 - 8}}{4} =$ $=(-6 +- \sqrt(36-8))/(4)=$
$= \frac{- 6 \pm \sqrt{28}}{4} =$ $=(-6 +- \sqrt(28))/(4)=$
$= \frac{- 6 \pm \sqrt{2^{2} \cdot 7}}{4} =$ $=(-6 +- \sqrt(2^2*7))/(4)=$
$= \frac{- 6 \pm 2 \cdot \sqrt{7}}{4} =$ $=(-6 +- 2* \sqrt(7))/(4)=$
$= \frac{- 3 \pm \sqrt{7}}{2}$ $=(-3 +- \sqrt(7))/(2)$

$x_{1} = \frac{- 3 + \sqrt{7}}{2}$ $x_1=(-3 + \sqrt(7))/(2)$
$x_{2} = \frac{- 3 - \sqrt{7}}{2}$ $x_2=(-3 - \sqrt(7))/(2)$

Answer 3 · 2018-05-08T13:59:26

$x = \frac{- 3 + \sqrt{7}}{2} or x = \frac{- 3 - \sqrt{7}}{2}$ $x=(-3+sqrt7)/2 or x=(-3-sqrt7)/2$

Explanation:

Here,

$2 x^{2} + 3 x - 3 = - 3 x - 4$ $2x^2+3x-3=-3x-4$

$\Rightarrow 2 x^{2} + 3 x - 3 + 3 x + 4 = 0$ $=>2x^2+3x-3+3x+4=0$

$\Rightarrow 2 x^{2} + 6 x + 1 = 0$ $=>2x^2+6x+1=0$

Comparing with $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ ,

$a = 2, b = 6 and c = 1$ $a=2,b=6 and c=1$

So,

$△ = b^{2} - 4 a c = 36 - 4 (2) (1) = 36 - 8 = 28$ $triangle=b^2-4ac=36-4(2)(1)=36-8=28$

$\sqrt{△} = 2 \sqrt{7}$ $sqrt(triangle)=2sqrt7$

$:.x=(-b+-sqrt(triangle))/(2a)=(-6+-2sqrt7)/(2(2))=(-3+-sqrt7)/2$

Hence,

$x=(-3+sqrt7)/2 or x=(-3-sqrt7)/2$

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How do you solve using the quadratic form? 2x^2+3x-3=-3x-4

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