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32x^2-28x+3=0
a = 32; b = -28; c = +3;
Δ = b2-4ac
Δ = -282-4·32·3
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-20}{2*32}=\frac{8}{64} =1/8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+20}{2*32}=\frac{48}{64} =3/4 $
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