-5(x+2)2;x=-3

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Solution for -5(x+2)2;x=-3 equation:



-5(x+2)2x=-3
We move all terms to the left:
-5(x+2)2x-(-3)=0
We add all the numbers together, and all the variables
-5(x+2)2x+3=0
We multiply parentheses
-10x^2-20x+3=0
a = -10; b = -20; c = +3;
Δ = b2-4ac
Δ = -202-4·(-10)·3
Δ = 520
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{130}}{2*-10}=\frac{20-2\sqrt{130}}{-20} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{130}}{2*-10}=\frac{20+2\sqrt{130}}{-20} $

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