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

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

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

We move all terms to the left:

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

We multiply parentheses

-5x^2+4x+15x-(3(x+1)-4)=0


We calculate terms in parentheses: -(3(x+1)-4), so:

3(x+1)-4

We multiply parentheses

3x+3-4

We add all the numbers together, and all the variables

3x-1

Back to the equation:

-(3x-1)


We add all the numbers together, and all the variables

-5x^2+19x-(3x-1)=0

We get rid of parentheses

-5x^2+19x-3x+1=0

We add all the numbers together, and all the variables

-5x^2+16x+1=0

a = -5; b = 16; c = +1;
Δ = b2-4ac
Δ = 162-4·(-5)·1
Δ = 276
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{69}}{2*-5}=\frac{-16-2\sqrt{69}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{69}}{2*-5}=\frac{-16+2\sqrt{69}}{-10}
$