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360+2x^2-94x=0
a = 2; b = -94; c = +360;
Δ = b2-4ac
Δ = -942-4·2·360
Δ = 5956
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5956}=\sqrt{4*1489}=\sqrt{4}*\sqrt{1489}=2\sqrt{1489}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-94)-2\sqrt{1489}}{2*2}=\frac{94-2\sqrt{1489}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-94)+2\sqrt{1489}}{2*2}=\frac{94+2\sqrt{1489}}{4} $
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