4t^2+8t-9=0

Solution for 4t^2+8t-9=0 equation:

4t^2+8t-9=0

a = 4; b = 8; c = -9;
Δ = b2-4ac
Δ = 82-4·4·(-9)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{13}}{2*4}=\frac{-8-4\sqrt{13}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{13}}{2*4}=\frac{-8+4\sqrt{13}}{8}
$