2x^2+3x+7x=21

Solution for 2x^2+3x+7x=21 equation:

2x^2+3x+7x=21

We move all terms to the left:

2x^2+3x+7x-(21)=0

We add all the numbers together, and all the variables

2x^2+10x-21=0

a = 2; b = 10; c = -21;
Δ = b2-4ac
Δ = 102-4·2·(-21)
Δ = 268
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:

$\sqrt{\Delta}=\sqrt{268}=\sqrt{4*67}=\sqrt{4}*\sqrt{67}=2\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{67}}{2*2}=\frac{-10-2\sqrt{67}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{67}}{2*2}=\frac{-10+2\sqrt{67}}{4}
$