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3x^2-32x+64=0
a = 3; b = -32; c = +64;
Δ = b2-4ac
Δ = -322-4·3·64
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16}{2*3}=\frac{16}{6} =2+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16}{2*3}=\frac{48}{6} =8 $
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