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2x^2+18x+24=0
a = 2; b = 18; c = +24;
Δ = b2-4ac
Δ = 182-4·2·24
Δ = 132
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{33}}{2*2}=\frac{-18-2\sqrt{33}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{33}}{2*2}=\frac{-18+2\sqrt{33}}{4} $
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