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-0.5x^2-45x-215=0
a = -0.5; b = -45; c = -215;
Δ = b2-4ac
Δ = -452-4·(-0.5)·(-215)
Δ = 1595
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-\sqrt{1595}}{2*-0.5}=\frac{45-\sqrt{1595}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+\sqrt{1595}}{2*-0.5}=\frac{45+\sqrt{1595}}{-1} $
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