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3x^2-15x-12=0
a = 3; b = -15; c = -12;
Δ = b2-4ac
Δ = -152-4·3·(-12)
Δ = 369
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{41}}{2*3}=\frac{15-3\sqrt{41}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{41}}{2*3}=\frac{15+3\sqrt{41}}{6} $
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