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3x^2+11x-32=0
a = 3; b = 11; c = -32;
Δ = b2-4ac
Δ = 112-4·3·(-32)
Δ = 505
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{505}}{2*3}=\frac{-11-\sqrt{505}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{505}}{2*3}=\frac{-11+\sqrt{505}}{6} $
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