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2c^2-5c-180=0
a = 2; b = -5; c = -180;
Δ = b2-4ac
Δ = -52-4·2·(-180)
Δ = 1465
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{1465}}{2*2}=\frac{5-\sqrt{1465}}{4} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{1465}}{2*2}=\frac{5+\sqrt{1465}}{4} $
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