How do you solve $x^{2} + 5 x = - 2$ $x^2+5x=-2$ using the quadratic formula?

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How do you solve $x^{2} + 5 x = - 2$ $x^2+5x=-2$ using the quadratic formula?

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Answer 1 · 2015-07-10T03:32:04

Solve $y = x^{2} + 5 x + 2 = 0$ $y = x^2 + 5x + 2 = 0$

Explanation:

$D = d^{2} = b^{2} - 4 a c = 25 - 8 = 16 = 4^{2} \to d = \pm 4$ $D = d^2 = b^2 - 4ac = 25 - 8 = 16 = 4^2 -> d = +- 4$

$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{5}{2} \pm \frac{4}{2}$ $x = -b/(2a) +- d/(2a) = - 5/2 +- 4/2$

Answer 2 · 2015-07-10T03:52:25

$x = \frac{- 5 + \sqrt{17}}{2},$ $x=(-5+sqrt(17))/2,$ $\frac{- 5 - \sqrt{17}}{2}$ $(-5-sqrt(17))/2$

Explanation:

$x^{2} + 5 x = - 2$ $x^2+5x=-2$

Quadratic Equation
Add $2$ $2$ to both sides of the equation in order to make it fit the form of a quadratic equation: $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ .

$x^{2} + 5 x + 2 = 0$ $x^2+5x+2=0$

$a = 1;$ $a=1;$ $b = 5;$ $b=5;$ $c = 2$ $c=2$

Quadratic Formula

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

$x = \frac{- 5 \pm \sqrt{5^{2} - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$ $x=(-5+-sqrt(5^2-4*1*2))/(2*1)$

$x = \frac{- 5 \pm \sqrt{25 - 8}}{2}$ $x=(-5+-sqrt(25-8))/2$ =

$x = \frac{- 5 \pm \sqrt{17}}{2}$ $x=(-5+-sqrt(17))/2$

$x = \frac{- 5 + \sqrt{17}}{2}$ $x=(-5+sqrt(17))/2$

$x = \frac{- 5 - \sqrt{17}}{2}$ $x=(-5-sqrt(17))/2$

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How do you solve $x^{2} + 5 x = - 2$ $x^2+5x=-2$ using the quadratic formula?

2 Answers

Explanation:

Explanation:

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How do you solve x2+5x=−2x^2+5x=-2 using the quadratic formula?

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How do you solve $x^{2} + 5 x = - 2$ $x^2+5x=-2$ using the quadratic formula?