3x^2+4x+7=)

Solution for 3x^2+4x+7=) equation:

3x^2+4x+7=)

We move all terms to the left:

3x^2+4x+7-())=0

We add all the numbers together, and all the variables

3x^2+4x=0

a = 3; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·3·0
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*3}=\frac{-8}{6}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*3}=\frac{0}{6}
=0
$