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126+2x^2-32x=0
a = 2; b = -32; c = +126;
Δ = b2-4ac
Δ = -322-4·2·126
Δ = 16
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4}{2*2}=\frac{28}{4} =7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4}{2*2}=\frac{36}{4} =9 $
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