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28=32x-4x^2
We move all terms to the left:
28-(32x-4x^2)=0
We get rid of parentheses
4x^2-32x+28=0
a = 4; b = -32; c = +28;
Δ = b2-4ac
Δ = -322-4·4·28
Δ = 576
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{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-24}{2*4}=\frac{8}{8} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+24}{2*4}=\frac{56}{8} =7 $
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