49x^2+28x-24=38x^2-4x

Solution for 49x^2+28x-24=38x^2-4x equation:

49x^2+28x-24=38x^2-4x

We move all terms to the left:

49x^2+28x-24-(38x^2-4x)=0

We get rid of parentheses

49x^2-38x^2+28x+4x-24=0

We add all the numbers together, and all the variables

11x^2+32x-24=0

a = 11; b = 32; c = -24;
Δ = b2-4ac
Δ = 322-4·11·(-24)
Δ = 2080
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:

$\sqrt{\Delta}=\sqrt{2080}=\sqrt{16*130}=\sqrt{16}*\sqrt{130}=4\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{130}}{2*11}=\frac{-32-4\sqrt{130}}{22}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{130}}{2*11}=\frac{-32+4\sqrt{130}}{22}
$