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20x^2+3x-180=0
a = 20; b = 3; c = -180;
Δ = b2-4ac
Δ = 32-4·20·(-180)
Δ = 14409
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{14409}=\sqrt{9*1601}=\sqrt{9}*\sqrt{1601}=3\sqrt{1601}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{1601}}{2*20}=\frac{-3-3\sqrt{1601}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{1601}}{2*20}=\frac{-3+3\sqrt{1601}}{40} $
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