(7p+6)(2p-5)=-2p(p+9)-19

Solution for (7p+6)(2p-5)=-2p(p+9)-19 equation:

(7p+6)(2p-5)=-2p(p+9)-19

We move all terms to the left:

(7p+6)(2p-5)-(-2p(p+9)-19)=0

We multiply parentheses ..

(+14p^2-35p+12p-30)-(-2p(p+9)-19)=0


We calculate terms in parentheses: -(-2p(p+9)-19), so:

-2p(p+9)-19

We multiply parentheses

-2p^2-18p-19

Back to the equation:

-(-2p^2-18p-19)


We get rid of parentheses

14p^2+2p^2-35p+12p+18p-30+19=0

We add all the numbers together, and all the variables

16p^2-5p-11=0

a = 16; b = -5; c = -11;
Δ = b2-4ac
Δ = -52-4·16·(-11)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-27}{2*16}=\frac{-22}{32}
=-11/16
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+27}{2*16}=\frac{32}{32}
=1
$