2u^2-9u-19=(u-3)(u-3)

Solution for 2u^2-9u-19=(u-3)(u-3) equation:

2u^2-9u-19=(u-3)(u-3)

We move all terms to the left:

2u^2-9u-19-((u-3)(u-3))=0

We multiply parentheses ..

2u^2-((+u^2-3u-3u+9))-9u-19=0


We calculate terms in parentheses: -((+u^2-3u-3u+9)), so:

(+u^2-3u-3u+9)

We get rid of parentheses

u^2-3u-3u+9

We add all the numbers together, and all the variables

u^2-6u+9

Back to the equation:

-(u^2-6u+9)


We add all the numbers together, and all the variables

2u^2-9u-(u^2-6u+9)-19=0

We get rid of parentheses

2u^2-u^2-9u+6u-9-19=0

We add all the numbers together, and all the variables

u^2-3u-28=0

a = 1; b = -3; c = -28;
Δ = b2-4ac
Δ = -32-4·1·(-28)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*1}=\frac{-8}{2}
=-4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*1}=\frac{14}{2}
=7
$