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-2a^2+32=0
a = -2; b = 0; c = +32;
Δ = b2-4ac
Δ = 02-4·(-2)·32
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-2}=\frac{-16}{-4} =+4 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-2}=\frac{16}{-4} =-4 $
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