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3x^2+95x-32=0
a = 3; b = 95; c = -32;
Δ = b2-4ac
Δ = 952-4·3·(-32)
Δ = 9409
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{9409}=97$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(95)-97}{2*3}=\frac{-192}{6} =-32 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(95)+97}{2*3}=\frac{2}{6} =1/3 $
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