3x(4x+5)=4(5x+12)+23=

Solution for 3x(4x+5)=4(5x+12)+23= equation:

3x(4x+5)=4(5x+12)+23=

We move all terms to the left:

3x(4x+5)-(4(5x+12)+23)=0

We multiply parentheses

12x^2+15x-(4(5x+12)+23)=0


We calculate terms in parentheses: -(4(5x+12)+23), so:

4(5x+12)+23

We multiply parentheses

20x+48+23

We add all the numbers together, and all the variables

20x+71

Back to the equation:

-(20x+71)


We get rid of parentheses

12x^2+15x-20x-71=0

We add all the numbers together, and all the variables

12x^2-5x-71=0

a = 12; b = -5; c = -71;
Δ = b2-4ac
Δ = -52-4·12·(-71)
Δ = 3433
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{-(-5)-\sqrt{3433}}{2*12}=\frac{5-\sqrt{3433}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{3433}}{2*12}=\frac{5+\sqrt{3433}}{24}
$