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2x^2+32x-175=0
a = 2; b = 32; c = -175;
Δ = b2-4ac
Δ = 322-4·2·(-175)
Δ = 2424
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2424}=\sqrt{4*606}=\sqrt{4}*\sqrt{606}=2\sqrt{606}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-2\sqrt{606}}{2*2}=\frac{-32-2\sqrt{606}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+2\sqrt{606}}{2*2}=\frac{-32+2\sqrt{606}}{4} $
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