How do you solve $2 x^{2} + 12 x = 10$ $2x^2 + 12x= 10$ using the quadratic formula?

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Answer 1 · 2017-05-09T16:14:24

$x = - 3 \pm \sqrt{14}$ $x=-3+-sqrt14$

first rearrange to the form $a x^{2} + b x + c = 0$ $" "ax^2+bx+c=0$

we have

$2 x^{2} + 12 x - 10 = 0$ $2x^2+12x-10=0$

divide out any common factors to make it simpler to use.

$(2 x^{2} + 12 x - 10 = 0) \div 2$ $(2x^2+12x-10=0)-:2$

$\Rightarrow x^{2} + 6 x - 5 = 0$ $=>x^2+6x-5=0$

the formula is

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

$a = 1, b = 6, c = - 5$ $a=1, b=6, c=-5$

$x = \frac{- 6 \pm \sqrt{6^{2} - 4 \times 1 \times (- 5)}}{2 \times 1}$ $x=(-6+-sqrt(6^2-4xx1xx(-5)))/(2xx1)$

$x = \frac{- 6 \pm \sqrt{36 + 20}}{2}$ $x=(-6+-sqrt(36+20))/2$

$x = \frac{- 6 \pm \sqrt{56}}{2}$ $x=(-6+-sqrt56)/2$

$x = - \frac{6}{2} \pm \frac{\sqrt{56}}{2}$ $x=-6/2+-sqrt56/2$

$x = - \frac{6}{2} \pm 2 \frac{\sqrt{14}}{2}$ $x=-6/2+-2sqrt14/2$

$:.x=-3+-sqrt14$

How do you solve 2x^2 + 12x= 10 2x2+12x=102x^2 + 12x= 10 using the quadratic formula?