4t^2+13t=-3

Solution for 4t^2+13t=-3 equation:

4t^2+13t=-3

We move all terms to the left:

4t^2+13t-(-3)=0

We add all the numbers together, and all the variables

4t^2+13t+3=0

a = 4; b = 13; c = +3;
Δ = b2-4ac
Δ = 132-4·4·3
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*4}=\frac{-24}{8}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*4}=\frac{-2}{8}
=-1/4
$