8t^2-9=0

Solution for 8t^2-9=0 equation:

8t^2-9=0

a = 8; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·8·(-9)
Δ = 288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*8}=\frac{0-12\sqrt{2}}{16}
=-\frac{12\sqrt{2}}{16}
=-\frac{3\sqrt{2}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*8}=\frac{0+12\sqrt{2}}{16}
=\frac{12\sqrt{2}}{16}
=\frac{3\sqrt{2}}{4}
$