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32x^2-18=0
a = 32; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·32·(-18)
Δ = 2304
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{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*32}=\frac{-48}{64} =-3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*32}=\frac{48}{64} =3/4 $
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