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2x^2+12x-432=0
a = 2; b = 12; c = -432;
Δ = b2-4ac
Δ = 122-4·2·(-432)
Δ = 3600
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{3600}=60$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-60}{2*2}=\frac{-72}{4} =-18 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+60}{2*2}=\frac{48}{4} =12 $
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