7z^2+7z-2=(8z+1)(z-2)

Solution for 7z^2+7z-2=(8z+1)(z-2) equation:

7z^2+7z-2=(8z+1)(z-2)

We move all terms to the left:

7z^2+7z-2-((8z+1)(z-2))=0

We multiply parentheses ..

7z^2-((+8z^2-16z+z-2))+7z-2=0


We calculate terms in parentheses: -((+8z^2-16z+z-2)), so:

(+8z^2-16z+z-2)

We get rid of parentheses

8z^2-16z+z-2

We add all the numbers together, and all the variables

8z^2-15z-2

Back to the equation:

-(8z^2-15z-2)


We add all the numbers together, and all the variables

7z^2+7z-(8z^2-15z-2)-2=0

We get rid of parentheses

7z^2-8z^2+7z+15z+2-2=0

We add all the numbers together, and all the variables

-1z^2+22z=0

a = -1; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·(-1)·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*-1}=\frac{-44}{-2}
=+22
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*-1}=\frac{0}{-2}
=0
$