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2x^2-28x+192=96
We move all terms to the left:
2x^2-28x+192-(96)=0
We add all the numbers together, and all the variables
2x^2-28x+96=0
a = 2; b = -28; c = +96;
Δ = b2-4ac
Δ = -282-4·2·96
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4}{2*2}=\frac{24}{4} =6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4}{2*2}=\frac{32}{4} =8 $
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