2x^2+9x+4=x+4(21)

Solution for 2x^2+9x+4=x+4(21) equation:

2x^2+9x+4=x+4(21)

We move all terms to the left:

2x^2+9x+4-(x+4(21))=0

We get rid of parentheses

2x^2+9x-x-421+4=0

We add all the numbers together, and all the variables

2x^2+8x-417=0

a = 2; b = 8; c = -417;
Δ = b2-4ac
Δ = 82-4·2·(-417)
Δ = 3400
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:

$\sqrt{\Delta}=\sqrt{3400}=\sqrt{100*34}=\sqrt{100}*\sqrt{34}=10\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-10\sqrt{34}}{2*2}=\frac{-8-10\sqrt{34}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+10\sqrt{34}}{2*2}=\frac{-8+10\sqrt{34}}{4}
$