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2x^2+22x-63=0
a = 2; b = 22; c = -63;
Δ = b2-4ac
Δ = 222-4·2·(-63)
Δ = 988
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{988}=\sqrt{4*247}=\sqrt{4}*\sqrt{247}=2\sqrt{247}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{247}}{2*2}=\frac{-22-2\sqrt{247}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{247}}{2*2}=\frac{-22+2\sqrt{247}}{4} $
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