Solve 5-3x^2/6-x=2-3x/4

$$\frac { 5 - 3 x ^ { 2 } } { 6 } - x = \frac { 2 - 3 x } { 4 }$$

Solve for x

$$x=\frac{\sqrt{105}}{12}-\frac{1}{4}\approx 0.603912564$$

$$x=-\frac{\sqrt{105}}{12}-\frac{1}{4}\approx -1.103912564$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 12, the least common multiple of 6,4.

$$2\left(5-3x^{2}\right)-12x=3\left(2-3x\right)$$

Use the distributive property to multiply 2 by 5-3x^{2}.

$$10-6x^{2}-12x=3\left(2-3x\right)$$

Use the distributive property to multiply 3 by 2-3x.

$$10-6x^{2}-12x=6-9x$$

Subtract 6 from both sides.

$$10-6x^{2}-12x-6=-9x$$

Subtract 6 from 10 to get 4.

$$4-6x^{2}-12x=-9x$$

Add 9x to both sides.

$$4-6x^{2}-12x+9x=0$$

Combine -12x and 9x to get -3x.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-6\right)\times 4}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-3\right)±\sqrt{9+24\times 4}}{2\left(-6\right)}$$

Multiply 24 times 4.

$$x=\frac{-\left(-3\right)±\sqrt{9+96}}{2\left(-6\right)}$$

Add 9 to 96.

$$x=\frac{-\left(-3\right)±\sqrt{105}}{2\left(-6\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{105}}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{3±\sqrt{105}}{-12}$$

Now solve the equation x=\frac{3±\sqrt{105}}{-12} when ± is plus. Add 3 to \sqrt{105}.

$$x=\frac{\sqrt{105}+3}{-12}$$

Divide 3+\sqrt{105} by -12.

$$x=-\frac{\sqrt{105}}{12}-\frac{1}{4}$$

Now solve the equation x=\frac{3±\sqrt{105}}{-12} when ± is minus. Subtract \sqrt{105} from 3.

$$x=\frac{3-\sqrt{105}}{-12}$$

Divide 3-\sqrt{105} by -12.

$$x=\frac{\sqrt{105}}{12}-\frac{1}{4}$$

The equation is now solved.

$$x=-\frac{\sqrt{105}}{12}-\frac{1}{4} x=\frac{\sqrt{105}}{12}-\frac{1}{4}$$

Steps for Completing the Square

Multiply both sides of the equation by 12, the least common multiple of 6,4.

$$2\left(5-3x^{2}\right)-12x=3\left(2-3x\right)$$

Use the distributive property to multiply 2 by 5-3x^{2}.

$$10-6x^{2}-12x=3\left(2-3x\right)$$

Use the distributive property to multiply 3 by 2-3x.

$$10-6x^{2}-12x=6-9x$$

Add 9x to both sides.

$$10-6x^{2}-12x+9x=6$$

Combine -12x and 9x to get -3x.

$$10-6x^{2}-3x=6$$

Subtract 10 from both sides.

$$-6x^{2}-3x=6-10$$

Subtract 10 from 6 to get -4.

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

Divide both sides by -6.

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

Dividing by -6 undoes the multiplication by -6.

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

Reduce the fraction \frac{-3}{-6} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{1}{2}x=-\frac{4}{-6}$$

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

$$x^{2}+\frac{1}{2}x=\frac{2}{3}$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{2}{3}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{2}{3}+\frac{1}{16}$$

Add \frac{2}{3} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{35}{48}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{35}{48}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{35}{48}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{105}}{12} x+\frac{1}{4}=-\frac{\sqrt{105}}{12}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{\sqrt{105}}{12}-\frac{1}{4} x=-\frac{\sqrt{105}}{12}-\frac{1}{4}$$