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5x^2=-4x+4
We move all terms to the left:
5x^2-(-4x+4)=0
We get rid of parentheses
5x^2+4x-4=0
a = 5; b = 4; c = -4;
Δ = b2-4ac
Δ = 42-4·5·(-4)
Δ = 96
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{6}}{2*5}=\frac{-4-4\sqrt{6}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{6}}{2*5}=\frac{-4+4\sqrt{6}}{10} $
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