How do you solve $2 x^{2} + 10 x + 12 = 0$ $2x^2 + 10x + 12 = 0$ using the quadratic formula?

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How do you solve $2 x^{2} + 10 x + 12 = 0$ $2x^2 + 10x + 12 = 0$ using the quadratic formula?

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Answer 1 · 2016-03-15T19:41:50

$x_{1} = - 3$ $x_1=-3$
$x_{2} = - 2$ $x_2=-2$

Explanation:

$2 x^{2} + 10 x + 12 = 0$ $2x^2+10x+12=0$
$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

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

$x_{1} = \frac{- 10 - \sqrt{100 - 4 \cdot 2 \cdot 12}}{2 \cdot 2}$ $x_1=(-10-sqrt(100-4*2*12))/(2*2)$
$x_{1} = \frac{- 10 - \sqrt{4}}{4}$ $x_1=(-10-sqrt4)/4$
$x_{1} = \frac{- 10 - 2}{4}$ $x_1=(-10-2)/4$
$x_{1} = - \frac{12}{4} x_{1} = - 3$ $x_1=-12/4" "x_1=-3$

$x_{2} = \frac{- 10 + \sqrt{100 - 4 \cdot 2 \cdot 12}}{2 \cdot 2}$ $x_2=(-10+sqrt(100-4*2*12))/(2*2)$
$x_{2} = \frac{- 10 + 2}{4}$ $x_2=(-10+2)/4$

$x_{2} = - \frac{8}{4}$ $x_2=-8/4$
$x_{2} = - 2$ $x_2=-2$

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How do you solve $2 x^{2} + 10 x + 12 = 0$ $2x^2 + 10x + 12 = 0$ using the quadratic formula?

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Explanation:

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