3x+4+7x^2=199

Solution for 3x+4+7x^2=199 equation:

3x+4+7x^2=199

We move all terms to the left:

3x+4+7x^2-(199)=0

We add all the numbers together, and all the variables

7x^2+3x-195=0

a = 7; b = 3; c = -195;
Δ = b2-4ac
Δ = 32-4·7·(-195)
Δ = 5469
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{5469}}{2*7}=\frac{-3-\sqrt{5469}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{5469}}{2*7}=\frac{-3+\sqrt{5469}}{14}
$