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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -(-4(2x+1)), so:

-4(2x+1)

We multiply parentheses

-8x-4

Back to the equation:

-(-8x-4)


We get rid of parentheses

3x^2+8x+4=0

a = 3; b = 8; c = +4;
Δ = b2-4ac
Δ = 82-4·3·4
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{-(8)-4}{2*3}=\frac{-12}{6}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*3}=\frac{-4}{6}
=-2/3
$