(4z+5)(z+1)=4^2+131

Solution for (4z+5)(z+1)=4^2+131 equation:

(4z+5)(z+1)=4^2+131

We move all terms to the left:

(4z+5)(z+1)-(4^2+131)=0

We add all the numbers together, and all the variables

(4z+5)(z+1)-147=0

We multiply parentheses ..

(+4z^2+4z+5z+5)-147=0

We get rid of parentheses

4z^2+4z+5z+5-147=0

We add all the numbers together, and all the variables

4z^2+9z-142=0

a = 4; b = 9; c = -142;
Δ = b2-4ac
Δ = 92-4·4·(-142)
Δ = 2353
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{2353}}{2*4}=\frac{-9-\sqrt{2353}}{8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{2353}}{2*4}=\frac{-9+\sqrt{2353}}{8}
$