Question

Solve each equation by factoring or by using the quadratic formula. The quadratic formula is:

If$a{x}^2+bx+c=0$, with$a\ne 0$, then $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

$3x^2+3x=4$

Short answer

The values of x are$x=\frac{-3+\sqrt{57}}{6}$ and$x=\frac{-3-\sqrt{57}}{6}$.

Step-by-step solution

- Understand the concept used 

If$a{x}^2+bx+c=0$ , with$a\ne 0$, then . Solving using factoring can be done by splitting the middle term of equation and then making groups and equating them to zero.$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

- Substitute the values 

It is given that $3x^2+3x=4$which can also be written as$3x^2+3x-4=0$. Using quadratic formula$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , substitute 3 for a, 3 for b and -4 for c.

Put the values in the formula and get $x=\frac{-3\pm \sqrt{{\left(3\right)}^2-4\times \left(3\right)\times \left(-4\right)}}{2\left(3\right)}$.

- Simplify for x 

Simplify the expression for x.

$\begin{array}{c}x=\frac{-3\pm \sqrt{{\left(3\right)}^2-4\times \left(3\right)\times \left(-4\right)}}{2\left(3\right)}\\ =\frac{-3\pm \sqrt{9+48}}{6}\\ =\frac{-3\pm \sqrt{57}}{6}\end{array}$

Therefore,$x=\frac{-3+\sqrt{57}}{6}$ and $x=\frac{-3-\sqrt{57}}{6}$.