3x^2+26x+19=-4x^2+4

Solution for 3x^2+26x+19=-4x^2+4 equation:

3x^2+26x+19=-4x^2+4

We move all terms to the left:

3x^2+26x+19-(-4x^2+4)=0

We get rid of parentheses

3x^2+4x^2+26x-4+19=0

We add all the numbers together, and all the variables

7x^2+26x+15=0

a = 7; b = 26; c = +15;
Δ = b2-4ac
Δ = 262-4·7·15
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-16}{2*7}=\frac{-42}{14}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+16}{2*7}=\frac{-10}{14}
=-5/7
$