(x+4)(3x-7)=-(x+4)10

Solution for (x+4)(3x-7)=-(x+4)10 equation:

(x+4)(3x-7)=-(x+4)10

We move all terms to the left:

(x+4)(3x-7)-(-(x+4)10)=0

We multiply parentheses ..

(+3x^2-7x+12x-28)-(-(x+4)10)=0


We calculate terms in parentheses: -(-(x+4)10), so:

-(x+4)10

We multiply parentheses

-10x-40

Back to the equation:

-(-10x-40)


We get rid of parentheses

3x^2-7x+12x+10x-28+40=0

We add all the numbers together, and all the variables

3x^2+15x+12=0

a = 3; b = 15; c = +12;
Δ = b2-4ac
Δ = 152-4·3·12
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-9}{2*3}=\frac{-24}{6}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+9}{2*3}=\frac{-6}{6}
=-1
$