(4x-3)(x+7)=(1-5x)(x-3)

Solution for (4x-3)(x+7)=(1-5x)(x-3) equation:

(4x-3)(x+7)=(1-5x)(x-3)

We move all terms to the left:

(4x-3)(x+7)-((1-5x)(x-3))=0

We add all the numbers together, and all the variables

(4x-3)(x+7)-((-5x+1)(x-3))=0

We multiply parentheses ..

(+4x^2+28x-3x-21)-((-5x+1)(x-3))=0


We calculate terms in parentheses: -((-5x+1)(x-3)), so:

(-5x+1)(x-3)

We multiply parentheses ..

(-5x^2+15x+x-3)

We get rid of parentheses

-5x^2+15x+x-3

We add all the numbers together, and all the variables

-5x^2+16x-3

Back to the equation:

-(-5x^2+16x-3)


We get rid of parentheses

4x^2+5x^2+28x-3x-16x-21+3=0

We add all the numbers together, and all the variables

9x^2+9x-18=0

a = 9; b = 9; c = -18;
Δ = b2-4ac
Δ = 92-4·9·(-18)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-27}{2*9}=\frac{-36}{18}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+27}{2*9}=\frac{18}{18}
=1
$