t^2=6t+4

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

t^2=6t+4

We move all terms to the left:

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

We get rid of parentheses

t^2-6t-4=0

a = 1; b = -6; c = -4;
Δ = b2-4ac
Δ = -62-4·1·(-4)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{13}}{2*1}=\frac{6-2\sqrt{13}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{13}}{2*1}=\frac{6+2\sqrt{13}}{2}
$