2x^2+3x+4x=3-7x

Solution for 2x^2+3x+4x=3-7x equation:

2x^2+3x+4x=3-7x

We move all terms to the left:

2x^2+3x+4x-(3-7x)=0

We add all the numbers together, and all the variables

2x^2+3x+4x-(-7x+3)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

2x^2+7x+7x-3=0

We add all the numbers together, and all the variables

2x^2+14x-3=0

a = 2; b = 14; c = -3;
Δ = b2-4ac
Δ = 142-4·2·(-3)
Δ = 220
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:

$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{55}}{2*2}=\frac{-14-2\sqrt{55}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{55}}{2*2}=\frac{-14+2\sqrt{55}}{4}
$