x2-(9x+19)=(x-5)

Solution for x2-(9x+19)=(x-5) equation:

x2-(9x+19)=(x-5)

We move all terms to the left:

x2-(9x+19)-((x-5))=0

We add all the numbers together, and all the variables

x^2-(9x+19)-((x-5))=0

We get rid of parentheses

x^2-9x-((x-5))-19=0


We calculate terms in parentheses: -((x-5)), so:

(x-5)

We get rid of parentheses

x-5

Back to the equation:

-(x-5)


We get rid of parentheses

x^2-9x-x+5-19=0

We add all the numbers together, and all the variables

x^2-10x-14=0

a = 1; b = -10; c = -14;
Δ = b2-4ac
Δ = -102-4·1·(-14)
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{39}}{2*1}=\frac{10-2\sqrt{39}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{39}}{2*1}=\frac{10+2\sqrt{39}}{2}
$