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3a^2+23a-8=0
a = 3; b = 23; c = -8;
Δ = b2-4ac
Δ = 232-4·3·(-8)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{625}=25$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-25}{2*3}=\frac{-48}{6} =-8 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+25}{2*3}=\frac{2}{6} =1/3 $
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