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15x^2+29x-12=0
a = 15; b = 29; c = -12;
Δ = b2-4ac
Δ = 292-4·15·(-12)
Δ = 1561
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{-(29)-\sqrt{1561}}{2*15}=\frac{-29-\sqrt{1561}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+\sqrt{1561}}{2*15}=\frac{-29+\sqrt{1561}}{30} $
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