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

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Answer 1 · 2015-07-10T03:32:04

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

$D = d^2 = b^2 - 4ac = 25 - 8 = 16 = 4^2 -> d = +- 4$

$x = -b/(2a) +- d/(2a) = - 5/2 +- 4/2$

Answer 2 · 2015-07-10T03:52:25

$x=(-5+sqrt(17))/2,$ $(-5-sqrt(17))/2$

$x^2+5x=-2$

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

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

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

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$x=(-5+-sqrt(5^2-4*1*2))/(2*1)$

$x=(-5+-sqrt(25-8))/2$ =

$x=(-5+-sqrt(17))/2$

$x=(-5+sqrt(17))/2$

$x=(-5-sqrt(17))/2$

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