(3x^2+1x+-4)=0

Solution for (3x^2+1x+-4)=0 equation:

(3x^2+1x+-4)=0

We use the square of the difference formula

(3x^2+1x-4)=0

We get rid of parentheses

3x^2+1x-4=0

We add all the numbers together, and all the variables

3x^2+x-4=0

a = 3; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·3·(-4)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*3}=\frac{-8}{6}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*3}=\frac{6}{6}
=1
$