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20x^2-3x-63=0
a = 20; b = -3; c = -63;
Δ = b2-4ac
Δ = -32-4·20·(-63)
Δ = 5049
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5049}=\sqrt{9*561}=\sqrt{9}*\sqrt{561}=3\sqrt{561}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{561}}{2*20}=\frac{3-3\sqrt{561}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{561}}{2*20}=\frac{3+3\sqrt{561}}{40} $
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