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40=(2x+2)x
We move all terms to the left:
40-((2x+2)x)=0
We calculate terms in parentheses: -((2x+2)x), so:We get rid of parentheses
(2x+2)x
We multiply parentheses
2x^2+2x
Back to the equation:
-(2x^2+2x)
-2x^2-2x+40=0
a = -2; b = -2; c = +40;
Δ = b2-4ac
Δ = -22-4·(-2)·40
Δ = 324
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-18}{2*-2}=\frac{-16}{-4} =+4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+18}{2*-2}=\frac{20}{-4} =-5 $
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