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5a^2+4a-57=0
a = 5; b = 4; c = -57;
Δ = b2-4ac
Δ = 42-4·5·(-57)
Δ = 1156
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{1156}=34$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-34}{2*5}=\frac{-38}{10} =-3+4/5 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+34}{2*5}=\frac{30}{10} =3 $
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