2x*(4+3x)-10=x^2-8

Solution for 2x*(4+3x)-10=x^2-8 equation:

2x(4+3x)-10=x^2-8

We move all terms to the left:

2x(4+3x)-10-(x^2-8)=0

We add all the numbers together, and all the variables

2x(3x+4)-(x^2-8)-10=0

We multiply parentheses

6x^2+8x-(x^2-8)-10=0

We get rid of parentheses

6x^2-x^2+8x+8-10=0

We add all the numbers together, and all the variables

5x^2+8x-2=0

a = 5; b = 8; c = -2;
Δ = b2-4ac
Δ = 82-4·5·(-2)
Δ = 104
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:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{26}}{2*5}=\frac{-8-2\sqrt{26}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{26}}{2*5}=\frac{-8+2\sqrt{26}}{10}
$