2u^2+21u+40=2u+35

Solution for 2u^2+21u+40=2u+35 equation:

2u^2+21u+40=2u+35

We move all terms to the left:

2u^2+21u+40-(2u+35)=0

We get rid of parentheses

2u^2+21u-2u-35+40=0

We add all the numbers together, and all the variables

2u^2+19u+5=0

a = 2; b = 19; c = +5;
Δ = b2-4ac
Δ = 192-4·2·5
Δ = 321
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{321}}{2*2}=\frac{-19-\sqrt{321}}{4}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{321}}{2*2}=\frac{-19+\sqrt{321}}{4}
$