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3x^2-32x-35=0
a = 3; b = -32; c = -35;
Δ = b2-4ac
Δ = -322-4·3·(-35)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-38}{2*3}=\frac{-6}{6} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+38}{2*3}=\frac{70}{6} =11+2/3 $
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