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0.5x^2+2x-90=0
a = 0.5; b = 2; c = -90;
Δ = b2-4ac
Δ = 22-4·0.5·(-90)
Δ = 184
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{46}}{2*0.5}=\frac{-2-2\sqrt{46}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{46}}{2*0.5}=\frac{-2+2\sqrt{46}}{1} $
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