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3x^2+4x-95=0
a = 3; b = 4; c = -95;
Δ = b2-4ac
Δ = 42-4·3·(-95)
Δ = 1156
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{1156}=34$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-34}{2*3}=\frac{-38}{6} =-6+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+34}{2*3}=\frac{30}{6} =5 $
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