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2x^2+38x-432=0
a = 2; b = 38; c = -432;
Δ = b2-4ac
Δ = 382-4·2·(-432)
Δ = 4900
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{4900}=70$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-70}{2*2}=\frac{-108}{4} =-27 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+70}{2*2}=\frac{32}{4} =8 $
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