Question

can someone explain the quadratic formula?

Answer 1

YOu want to prove the formula, or what to know how to apply it?

Answer 2

how do i apply it

Answer 3

when the equation is in the form: \(\large\color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }\).

Then, the solution for x is:
\(\color{ slate }{{\bbox[5pt, lightcyan ,border:2px solid black ]{ \LARGE{x=~} \huge{ \frac{-\color{magenta}{b} \pm\sqrt{ \color{magenta}{~b} ^2-4 \color{blue}{a} \color{red}{c}~}}{2 \color{blue}{a}} }~ }}}\)

Answer 4

For example, if you have:
\(\color{black}{ \color{blue}{3} x^2+ \color{magenta}{12}x+ \color{red}{7}=0 }\)

Then, the solution for x is:
\(\color{black}{ \displaystyle x=\frac{-\color{magenta}{12}\pm\sqrt{\color{magenta}{12}^2-4\cdot\color{blue}{3}\cdot \color{red}{7} }}{2\cdot\color{blue}{3} } }\)

\(\color{black}{ \displaystyle x=\frac{-\color{magenta}{12}\pm\sqrt{144-84 }}{6 }=\frac{-\color{magenta}{12}}{6 } \pm\frac{\sqrt{60 }}{6 } =-2 \pm\frac{2\sqrt{15 }}{6 } }\)

\(\color{black}{ \displaystyle x=-2 \pm\frac{\sqrt{15 }}{6} }\)