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(2x+2)(x-4)=-5
We move all terms to the left:
(2x+2)(x-4)-(-5)=0
We add all the numbers together, and all the variables
(2x+2)(x-4)+5=0
We multiply parentheses ..
(+2x^2-8x+2x-8)+5=0
We get rid of parentheses
2x^2-8x+2x-8+5=0
We add all the numbers together, and all the variables
2x^2-6x-3=0
a = 2; b = -6; c = -3;
Δ = b2-4ac
Δ = -62-4·2·(-3)
Δ = 60
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{15}}{2*2}=\frac{6-2\sqrt{15}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{15}}{2*2}=\frac{6+2\sqrt{15}}{4} $
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