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2x^2-3x-20=0
a = 2; b = -3; c = -20;
Δ = b2-4ac
Δ = -32-4·2·(-20)
Δ = 169
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*2}=\frac{-10}{4} =-2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*2}=\frac{16}{4} =4 $
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