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3x^2+15x-42=0
a = 3; b = 15; c = -42;
Δ = b2-4ac
Δ = 152-4·3·(-42)
Δ = 729
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{729}=27$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-27}{2*3}=\frac{-42}{6} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+27}{2*3}=\frac{12}{6} =2 $
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