4t^2=8t+6

Solution for 4t^2=8t+6 equation:

4t^2=8t+6

We move all terms to the left:

4t^2-(8t+6)=0

We get rid of parentheses

4t^2-8t-6=0

a = 4; b = -8; c = -6;
Δ = b2-4ac
Δ = -82-4·4·(-6)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{10}}{2*4}=\frac{8-4\sqrt{10}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{10}}{2*4}=\frac{8+4\sqrt{10}}{8}
$