x^2-3x+4+4x^2-x-12=

Solution for x^2-3x+4+4x^2-x-12= equation:

x^2-3x+4+4x^2-x-12=

We move all terms to the left:

x^2-3x+4+4x^2-x-12-()=0

We add all the numbers together, and all the variables

5x^2-4x=0

a = 5; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·5·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*5}=\frac{0}{10}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*5}=\frac{8}{10}
=4/5
$