3(x-4)2+5(x-3)2=(2x-5)(4x-1)-40

Solution for 3(x-4)2+5(x-3)2=(2x-5)(4x-1)-40 equation:

3(x-4)2+5(x-3)2=(2x-5)(4x-1)-40

We move all terms to the left:

3(x-4)2+5(x-3)2-((2x-5)(4x-1)-40)=0

We multiply parentheses

6x+10x-((2x-5)(4x-1)-40)-24-30=0

We multiply parentheses ..

-((+8x^2-2x-20x+5)-40)+6x+10x-24-30=0


We calculate terms in parentheses: -((+8x^2-2x-20x+5)-40), so:

(+8x^2-2x-20x+5)-40

We get rid of parentheses

8x^2-2x-20x+5-40

We add all the numbers together, and all the variables

8x^2-22x-35

Back to the equation:

-(8x^2-22x-35)


We add all the numbers together, and all the variables

16x-(8x^2-22x-35)-54=0

We get rid of parentheses

-8x^2+16x+22x+35-54=0

We add all the numbers together, and all the variables

-8x^2+38x-19=0

a = -8; b = 38; c = -19;
Δ = b2-4ac
Δ = 382-4·(-8)·(-19)
Δ = 836
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:

$\sqrt{\Delta}=\sqrt{836}=\sqrt{4*209}=\sqrt{4}*\sqrt{209}=2\sqrt{209}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{209}}{2*-8}=\frac{-38-2\sqrt{209}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{209}}{2*-8}=\frac{-38+2\sqrt{209}}{-16}
$