5(2x-1)(3x+4)=4(x+3)-27

Solution for 5(2x-1)(3x+4)=4(x+3)-27 equation:

5(2x-1)(3x+4)=4(x+3)-27

We move all terms to the left:

5(2x-1)(3x+4)-(4(x+3)-27)=0

We multiply parentheses ..

5(+6x^2+8x-3x-4)-(4(x+3)-27)=0


We calculate terms in parentheses: -(4(x+3)-27), so:

4(x+3)-27

We multiply parentheses

4x+12-27

We add all the numbers together, and all the variables

4x-15

Back to the equation:

-(4x-15)


We multiply parentheses

30x^2+40x-15x-(4x-15)-20=0

We get rid of parentheses

30x^2+40x-15x-4x+15-20=0

We add all the numbers together, and all the variables

30x^2+21x-5=0

a = 30; b = 21; c = -5;
Δ = b2-4ac
Δ = 212-4·30·(-5)
Δ = 1041
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{-(21)-\sqrt{1041}}{2*30}=\frac{-21-\sqrt{1041}}{60}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{1041}}{2*30}=\frac{-21+\sqrt{1041}}{60}
$