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2x^2+15x-360=0
a = 2; b = 15; c = -360;
Δ = b2-4ac
Δ = 152-4·2·(-360)
Δ = 3105
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3105}=\sqrt{9*345}=\sqrt{9}*\sqrt{345}=3\sqrt{345}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{345}}{2*2}=\frac{-15-3\sqrt{345}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{345}}{2*2}=\frac{-15+3\sqrt{345}}{4} $
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