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3x^2+20x-87=0
a = 3; b = 20; c = -87;
Δ = b2-4ac
Δ = 202-4·3·(-87)
Δ = 1444
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{1444}=38$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-38}{2*3}=\frac{-58}{6} =-9+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+38}{2*3}=\frac{18}{6} =3 $
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