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12x^2+15x-5=0
a = 12; b = 15; c = -5;
Δ = b2-4ac
Δ = 152-4·12·(-5)
Δ = 465
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{-(15)-\sqrt{465}}{2*12}=\frac{-15-\sqrt{465}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{465}}{2*12}=\frac{-15+\sqrt{465}}{24} $
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