Solve for b x=(-b+ square root of b^2-4ac)/(2a)

$x = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}$

Step 1

Rewrite the equation as $\frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} = x$ .

$\frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} = x$

Step 2

Simplify both sides of the equation.

Factor $- 1$ out of $- b$ .

$\frac{- (b) + \sqrt{b^{2} - 4 a c}}{2 a} = x$

Factor $- 1$ out of $\sqrt{b^{2} - 4 a c}$ .

$\frac{- (b) - 1 (- \sqrt{b^{2} - 4 a c})}{2 a} = x$

Factor $- 1$ out of $- (b) - 1 (- \sqrt{b^{2} - 4 a c})$ .

$\frac{- (b - \sqrt{b^{2} - 4 a c})}{2 a} = x$

Rewrite $- (b - \sqrt{b^{2} - 4 a c})$ as $- 1 (b - \sqrt{b^{2} - 4 a c})$ .

$\frac{- 1 (b - \sqrt{b^{2} - 4 a c})}{2 a} = x$

Move the negative in front of the fraction.

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

Step 3

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{b^{2} - 4 a c}$ as ${(b^{2} - 4 a c)}^{\frac{1}{2}}$ .

$- \frac{b - {(b^{2} - 4 a c)}^{\frac{1}{2}}}{2 a} = x$