(3x+4)(5x-7)=(2x+7)2+53

Solution for (3x+4)(5x-7)=(2x+7)2+53 equation:

(3x+4)(5x-7)=(2x+7)2+53

We move all terms to the left:

(3x+4)(5x-7)-((2x+7)2+53)=0

We multiply parentheses ..

(+15x^2-21x+20x-28)-((2x+7)2+53)=0


We calculate terms in parentheses: -((2x+7)2+53), so:

(2x+7)2+53

We multiply parentheses

4x+14+53

We add all the numbers together, and all the variables

4x+67

Back to the equation:

-(4x+67)


We get rid of parentheses

15x^2-21x+20x-4x-28-67=0

We add all the numbers together, and all the variables

15x^2-5x-95=0

a = 15; b = -5; c = -95;
Δ = b2-4ac
Δ = -52-4·15·(-95)
Δ = 5725
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:

$\sqrt{\Delta}=\sqrt{5725}=\sqrt{25*229}=\sqrt{25}*\sqrt{229}=5\sqrt{229}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5\sqrt{229}}{2*15}=\frac{5-5\sqrt{229}}{30}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5\sqrt{229}}{2*15}=\frac{5+5\sqrt{229}}{30}
$