Question

Solve the given quadratic equation by using the formula method, $(2x+3)(2x-2)+2=0$

Answer 1

The given equation $(2x+3)(2x-2)+2=0$ can be simplified as shown below:

$(2x+3)(3x-2)+2=0\Rightarrow \left[2x\right(3x-2)+3(3x-2\left)\right]+2=0\Rightarrow (6x^2-4x+9x-6)+2=0\Rightarrow 6x^2-4x+9x-6+2=0\Rightarrow 6x^2+5x-4=0$

Thequadratic equation $6x^2+5x-4=0$ is in the form $a{x}^2+bx+c=0$ where $a=6,b=5$ and $c=-4$.

We know that the quadratic formula is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, therefore, substitute $a=6,b=5$ and $c=-4$ in $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ as follows:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-5\pm \sqrt{5^2-(4\times 6\times -4)}}{2\times 6}=\frac{-5\pm \sqrt{25+96}}{12}=\frac{-5\pm \sqrt{121}}{12}=\frac{-5\pm 11}{12}$

$x=\frac{-5+11}{12}=\frac{6}{12}=\frac{1}{2},x=\frac{-5-11}{12}=-\frac{16}{12}=-\frac{4}{3}$

Hence, $x=\frac{1}{2}$ or $x=-\frac{4}{3}$.