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2x^2+24x-3=0
a = 2; b = 24; c = -3;
Δ = b2-4ac
Δ = 242-4·2·(-3)
Δ = 600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-10\sqrt{6}}{2*2}=\frac{-24-10\sqrt{6}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+10\sqrt{6}}{2*2}=\frac{-24+10\sqrt{6}}{4} $
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