4t^2=t+6

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

4t^2=t+6

We move all terms to the left:

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

We get rid of parentheses

4t^2-t-6=0

We add all the numbers together, and all the variables

4t^2-1t-6=0

a = 4; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·4·(-6)
Δ = 97
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{97}}{2*4}=\frac{1-\sqrt{97}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{97}}{2*4}=\frac{1+\sqrt{97}}{8}
$