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4x-4=-2x^2
We move all terms to the left:
4x-4-(-2x^2)=0
We get rid of parentheses
2x^2+4x-4=0
a = 2; b = 4; c = -4;
Δ = b2-4ac
Δ = 42-4·2·(-4)
Δ = 48
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{3}}{2*2}=\frac{-4-4\sqrt{3}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{3}}{2*2}=\frac{-4+4\sqrt{3}}{4} $
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