(4z+3)2+(5z)2=150Z=

Solution for (4z+3)2+(5z)2=150Z= equation:

(4z+3)2+(5z)2=150=

We move all terms to the left:

(4z+3)2+(5z)2-(150)=0

We add all the numbers together, and all the variables

5z^2+(4z+3)2-150=0

We multiply parentheses

5z^2+8z+6-150=0

We add all the numbers together, and all the variables

5z^2+8z-144=0

a = 5; b = 8; c = -144;
Δ = b2-4ac
Δ = 82-4·5·(-144)
Δ = 2944
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2944}=\sqrt{64*46}=\sqrt{64}*\sqrt{46}=8\sqrt{46}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{46}}{2*5}=\frac{-8-8\sqrt{46}}{10}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{46}}{2*5}=\frac{-8+8\sqrt{46}}{10}
$