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-21=15-2z^2
We move all terms to the left:
-21-(15-2z^2)=0
We get rid of parentheses
2z^2-15-21=0
We add all the numbers together, and all the variables
2z^2-36=0
a = 2; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·2·(-36)
Δ = 288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*2}=\frac{0-12\sqrt{2}}{4} =-\frac{12\sqrt{2}}{4} =-3\sqrt{2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*2}=\frac{0+12\sqrt{2}}{4} =\frac{12\sqrt{2}}{4} =3\sqrt{2} $
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