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8y^{2}=5
Add 5 to both sides. Anything plus zero gives itself.
y^{2}=\frac{5}{8}
Divide both sides by 8.
y=\frac{\sqrt{10}}{4} y=-\frac{\sqrt{10}}{4}
Take the square root of both sides of the equation.
8y^{2}-5=0
Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.
y=\frac{0±\sqrt{0^{2}-4\times 8\left(-5\right)}}{2\times 8}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 0 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{0±\sqrt{-4\times 8\left(-5\right)}}{2\times 8}
Square 0.
y=\frac{0±\sqrt{-32\left(-5\right)}}{2\times 8}
Multiply -4 times 8.
y=\frac{0±\sqrt{160}}{2\times 8}
Multiply -32 times -5.
y=\frac{0±4\sqrt{10}}{2\times 8}
Take the square root of 160.
y=\frac{0±4\sqrt{10}}{16}
Multiply 2 times 8.
y=\frac{\sqrt{10}}{4}
Now solve the equation y=\frac{0±4\sqrt{10}}{16} when ± is plus.
y=-\frac{\sqrt{10}}{4}
Now solve the equation y=\frac{0±4\sqrt{10}}{16} when ± is minus.
y=\frac{\sqrt{10}}{4} y=-\frac{\sqrt{10}}{4}
The equation is now solved.