3=t^2+7t

Solution for 3=t^2+7t equation:

3=t^2+7t

We move all terms to the left:

3-(t^2+7t)=0

We get rid of parentheses

-t^2-7t+3=0

We add all the numbers together, and all the variables

-1t^2-7t+3=0

a = -1; b = -7; c = +3;
Δ = b2-4ac
Δ = -72-4·(-1)·3
Δ = 61
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{-(-7)-\sqrt{61}}{2*-1}=\frac{7-\sqrt{61}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{61}}{2*-1}=\frac{7+\sqrt{61}}{-2}
$