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

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

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

We move all terms to the left:

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

We use the square of the difference formula

x^2+(3x+1^2)-(4x(2x-1))+25=0

We get rid of parentheses

x^2+3x-(4x(2x-1))+25+1^2=0


We calculate terms in parentheses: -(4x(2x-1)), so:

4x(2x-1)

We multiply parentheses

8x^2-4x

Back to the equation:

-(8x^2-4x)


We add all the numbers together, and all the variables

x^2+3x-(8x^2-4x)+26=0

We get rid of parentheses

x^2-8x^2+3x+4x+26=0

We add all the numbers together, and all the variables

-7x^2+7x+26=0

a = -7; b = 7; c = +26;
Δ = b2-4ac
Δ = 72-4·(-7)·26
Δ = 777
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{777}}{2*-7}=\frac{-7-\sqrt{777}}{-14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{777}}{2*-7}=\frac{-7+\sqrt{777}}{-14}
$