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2x^2+39x-42=0
a = 2; b = 39; c = -42;
Δ = b2-4ac
Δ = 392-4·2·(-42)
Δ = 1857
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{-(39)-\sqrt{1857}}{2*2}=\frac{-39-\sqrt{1857}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+\sqrt{1857}}{2*2}=\frac{-39+\sqrt{1857}}{4} $
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