P=2(4x^2-3x)+2(7x-2)

Solution for P=2(4x^2-3x)+2(7x-2) equation:

=2(4P^2-3P)+2(7P-2)

We move all terms to the left:

-(2(4P^2-3P)+2(7P-2))=0


We calculate terms in parentheses: -(2(4P^2-3P)+2(7P-2)), so:

2(4P^2-3P)+2(7P-2)

We multiply parentheses

8P^2-6P+14P-4

We add all the numbers together, and all the variables

8P^2+8P-4

Back to the equation:

-(8P^2+8P-4)


We get rid of parentheses

-8P^2-8P+4=0

a = -8; b = -8; c = +4;
Δ = b2-4ac
Δ = -82-4·(-8)·4
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{3}}{2*-8}=\frac{8-8\sqrt{3}}{-16}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{3}}{2*-8}=\frac{8+8\sqrt{3}}{-16}
$