Solve 4x^2=-6x+3

$$4 x ^ { 2 } = - 6 x + 3$$

Solve for x

$$x=\frac{\sqrt{21}-3}{4}\approx 0.395643924$$

$$x=\frac{-\sqrt{21}-3}{4}\approx -1.895643924$$

Steps Using the Quadratic Formula

Add 6x to both sides.

$$4x^{2}+6x=3$$

Subtract 3 from both sides.

$$4x^{2}+6x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 6 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 4\left(-3\right)}}{2\times 4}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 4\left(-3\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-6±\sqrt{36-16\left(-3\right)}}{2\times 4}$$

Multiply -16 times -3.

$$x=\frac{-6±\sqrt{36+48}}{2\times 4}$$

Add 36 to 48.

$$x=\frac{-6±\sqrt{84}}{2\times 4}$$

Take the square root of 84.

$$x=\frac{-6±2\sqrt{21}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-6±2\sqrt{21}}{8}$$

Now solve the equation x=\frac{-6±2\sqrt{21}}{8} when ± is plus. Add -6 to 2\sqrt{21}.

$$x=\frac{2\sqrt{21}-6}{8}$$

Divide -6+2\sqrt{21} by 8.

$$x=\frac{\sqrt{21}-3}{4}$$

Now solve the equation x=\frac{-6±2\sqrt{21}}{8} when ± is minus. Subtract 2\sqrt{21} from -6.

$$x=\frac{-2\sqrt{21}-6}{8}$$

Divide -6-2\sqrt{21} by 8.

$$x=\frac{-\sqrt{21}-3}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{21}-3}{4} x=\frac{-\sqrt{21}-3}{4}$$

Steps for Completing the Square

Add 6x to both sides.

$$4x^{2}+6x=3$$

Divide both sides by 4.

$$\frac{4x^{2}+6x}{4}=\frac{3}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{6}{4}x=\frac{3}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{3}{2}x=\frac{3}{4}$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=\frac{3}{4}+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{3}{4}+\frac{9}{16}$$

Add \frac{3}{4} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{21}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{21}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{21}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{21}}{4} x+\frac{3}{4}=-\frac{\sqrt{21}}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{\sqrt{21}-3}{4} x=\frac{-\sqrt{21}-3}{4}$$