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3x^2+24x-315=0
a = 3; b = 24; c = -315;
Δ = b2-4ac
Δ = 242-4·3·(-315)
Δ = 4356
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4356}=66$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-66}{2*3}=\frac{-90}{6} =-15 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+66}{2*3}=\frac{42}{6} =7 $
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