(5x)(x-3)+6=x(x-3)+11

Solution for (5x)(x-3)+6=x(x-3)+11 equation:

(5x)(x-3)+6=x(x-3)+11

We move all terms to the left:

(5x)(x-3)+6-(x(x-3)+11)=0

We multiply parentheses

5x^2-15x-(x(x-3)+11)+6=0


We calculate terms in parentheses: -(x(x-3)+11), so:

x(x-3)+11

We multiply parentheses

x^2-3x+11

Back to the equation:

-(x^2-3x+11)


We get rid of parentheses

5x^2-x^2-15x+3x-11+6=0

We add all the numbers together, and all the variables

4x^2-12x-5=0

a = 4; b = -12; c = -5;
Δ = b2-4ac
Δ = -122-4·4·(-5)
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{14}}{2*4}=\frac{12-4\sqrt{14}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{14}}{2*4}=\frac{12+4\sqrt{14}}{8}
$