solve the equation: 7x^2=2x-5

For the general quadratic equation a*x² + b*x +c = 0 the solutions are given by:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$  

By rearranging 7x² = 2x - 5 to 7x² - 2x + 5 = 0 we have here that a=7, b=-2, c=5 so

$$x=\frac{2\pm\sqrt{2^2-4*7*5}}{2*7}$$  which results in:

$${\mathtt{7}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{34}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{7}}}}\\ {\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{34}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{7}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{7}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.832\: \!993\: \!127\: \!834\: \!905\: \!8}}{i}\right)\\ {\mathtt{x}} = {\frac{{\mathtt{1}}}{{\mathtt{7}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.832\: \!993\: \!127\: \!834\: \!905\: \!8}}{i}\\ \end{array} \right\}$$

 plug in the equation=7x² + 2x - 5
then take it step by step=
7x² + 7x - 5x - 5

7x (x + 1) - 5(x + 1)

(x + 1)(7x - 5)


5q^4 - 28q² - 12

5q^4 - 30q² + 2q² - 12

5q² (q² - 6) + 2(q² - 6)

(q² - 6)(5q² + 2)




2x² - x - 10

2x² + 4x - 5x - 10

2x (x + 2) - 5(x + 2)

(x + 2)(2x - 5)



-36x² - 3x + 60

-3 (12x² + x - 20)

-3 (12x² + 16x - 15x - 20)

-3 [4x(3x + 4) - 5(3x + 4)]

-3 (3x + 4) (4x - 5)