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-(2x)^2-10x+10=0
a = -2; b = -10; c = +10;
Δ = b2-4ac
Δ = -102-4·(-2)·10
Δ = 180
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6\sqrt{5}}{2*-2}=\frac{10-6\sqrt{5}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6\sqrt{5}}{2*-2}=\frac{10+6\sqrt{5}}{-4} $
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