(-x+2)(5-25)=-2x(x-4)+7

Solution for (-x+2)(5-25)=-2x(x-4)+7 equation:

(-x+2)(5-25)=-2x(x-4)+7

We move all terms to the left:

(-x+2)(5-25)-(-2x(x-4)+7)=0

We add all the numbers together, and all the variables

(-1x+2)(-20)-(-2x(x-4)+7)=0

We multiply parentheses ..

(+20x-40)-(-2x(x-4)+7)=0


We calculate terms in parentheses: -(-2x(x-4)+7), so:

-2x(x-4)+7

We multiply parentheses

-2x^2+8x+7

Back to the equation:

-(-2x^2+8x+7)


We add all the numbers together, and all the variables

-(-2x^2+8x+7)+(20x-40)=0

We get rid of parentheses

2x^2-8x+20x-7-40=0

We add all the numbers together, and all the variables

2x^2+12x-47=0

a = 2; b = 12; c = -47;
Δ = b2-4ac
Δ = 122-4·2·(-47)
Δ = 520
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:

$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{130}}{2*2}=\frac{-12-2\sqrt{130}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{130}}{2*2}=\frac{-12+2\sqrt{130}}{4}
$