7p(3p-4)=2(2p-1)+10

Solution for 7p(3p-4)=2(2p-1)+10 equation:

7p(3p-4)=2(2p-1)+10

We move all terms to the left:

7p(3p-4)-(2(2p-1)+10)=0

We multiply parentheses

21p^2-28p-(2(2p-1)+10)=0


We calculate terms in parentheses: -(2(2p-1)+10), so:

2(2p-1)+10

We multiply parentheses

4p-2+10

We add all the numbers together, and all the variables

4p+8

Back to the equation:

-(4p+8)


We get rid of parentheses

21p^2-28p-4p-8=0

We add all the numbers together, and all the variables

21p^2-32p-8=0

a = 21; b = -32; c = -8;
Δ = b2-4ac
Δ = -322-4·21·(-8)
Δ = 1696
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:

$\sqrt{\Delta}=\sqrt{1696}=\sqrt{16*106}=\sqrt{16}*\sqrt{106}=4\sqrt{106}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{106}}{2*21}=\frac{32-4\sqrt{106}}{42}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{106}}{2*21}=\frac{32+4\sqrt{106}}{42}
$