1x2=3(x+4)

Solution for 1x2=3(x+4) equation:

1x^2=3(x+4)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(3(x+4)), so:

3(x+4)

We multiply parentheses

3x+12

Back to the equation:

-(3x+12)


We get rid of parentheses

x^2-3x-12=0

a = 1; b = -3; c = -12;
Δ = b2-4ac
Δ = -32-4·1·(-12)
Δ = 57
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{57}}{2*1}=\frac{3-\sqrt{57}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{57}}{2*1}=\frac{3+\sqrt{57}}{2}
$