3a^2+11a-16=4

Solution for 3a^2+11a-16=4 equation:

3a^2+11a-16=4

We move all terms to the left:

3a^2+11a-16-(4)=0

We add all the numbers together, and all the variables

3a^2+11a-20=0

a = 3; b = 11; c = -20;
Δ = b2-4ac
Δ = 112-4·3·(-20)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*3}=\frac{-30}{6}
=-5
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*3}=\frac{8}{6}
=1+1/3
$