2x^2-(x+5)=7(x+3)

Solution for 2x^2-(x+5)=7(x+3) equation:

2x^2-(x+5)=7(x+3)

We move all terms to the left:

2x^2-(x+5)-(7(x+3))=0

We get rid of parentheses

2x^2-x-(7(x+3))-5=0


We calculate terms in parentheses: -(7(x+3)), so:

7(x+3)

We multiply parentheses

7x+21

Back to the equation:

-(7x+21)


We add all the numbers together, and all the variables

2x^2-1x-(7x+21)-5=0

We get rid of parentheses

2x^2-1x-7x-21-5=0

We add all the numbers together, and all the variables

2x^2-8x-26=0

a = 2; b = -8; c = -26;
Δ = b2-4ac
Δ = -82-4·2·(-26)
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{17}}{2*2}=\frac{8-4\sqrt{17}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{17}}{2*2}=\frac{8+4\sqrt{17}}{4}
$