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-0.5x^2+36x-558=0
a = -0.5; b = 36; c = -558;
Δ = b2-4ac
Δ = 362-4·(-0.5)·(-558)
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{-(36)-6\sqrt{5}}{2*-0.5}=\frac{-36-6\sqrt{5}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+6\sqrt{5}}{2*-0.5}=\frac{-36+6\sqrt{5}}{-1} $
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