P(x)=(x-3)(2-5x)-4x+7

Solution for P(x)=(x-3)(2-5x)-4x+7 equation:

(P)=(P-3)(2-5P)-4P+7

We move all terms to the left:

(P)-((P-3)(2-5P)-4P+7)=0

We add all the numbers together, and all the variables

P-((P-3)(-5P+2)-4P+7)=0

We multiply parentheses ..

-((-5P^2+2P+15P-6)-4P+7)+P=0


We calculate terms in parentheses: -((-5P^2+2P+15P-6)-4P+7), so:

(-5P^2+2P+15P-6)-4P+7

We get rid of parentheses

-5P^2+2P+15P-4P-6+7

We add all the numbers together, and all the variables

-5P^2+13P+1

Back to the equation:

-(-5P^2+13P+1)


We get rid of parentheses

5P^2-13P+P-1=0

We add all the numbers together, and all the variables

5P^2-12P-1=0

a = 5; b = -12; c = -1;
Δ = b2-4ac
Δ = -122-4·5·(-1)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{41}}{2*5}=\frac{12-2\sqrt{41}}{10}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{41}}{2*5}=\frac{12+2\sqrt{41}}{10}
$