Solve 2x/3-(5/4-x^2)=0

$$\frac { 2 x } { 3 } - ( \frac { 5 } { 4 } - x ^ { 2 } ) = 0$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=\frac{5}{6}\approx 0.833333333$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 12, the least common multiple of 3,4.

$$4\times 2x-12\left(\frac{5}{4}-x^{2}\right)=0$$

Multiply 4 and 2 to get 8.

$$8x-12\left(\frac{5}{4}-x^{2}\right)=0$$

Use the distributive property to multiply -12 by \frac{5}{4}-x^{2}.

$$8x-15+12x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$12x^{2}+8x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 12x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=8 ab=12\left(-15\right)=-180$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -180.

$$-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15$$

Calculate the sum for each pair.

$$-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3$$

The solution is the pair that gives sum 8.

$$a=-10 b=18$$

Rewrite 12x^{2}+8x-15 as \left(12x^{2}-10x\right)+\left(18x-15\right).

$$\left(12x^{2}-10x\right)+\left(18x-15\right)$$

Factor out 2x in the first and 3 in the second group.

$$2x\left(6x-5\right)+3\left(6x-5\right)$$

Factor out common term 6x-5 by using distributive property.

$$\left(6x-5\right)\left(2x+3\right)$$

To find equation solutions, solve 6x-5=0 and 2x+3=0.

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by 12, the least common multiple of 3,4.

$$4\times 2x-12\left(\frac{5}{4}-x^{2}\right)=0$$

Multiply 4 and 2 to get 8.

$$8x-12\left(\frac{5}{4}-x^{2}\right)=0$$

Use the distributive property to multiply -12 by \frac{5}{4}-x^{2}.

$$8x-15+12x^{2}=0$$

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.

$$12x^{2}+8x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 8 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 12\left(-15\right)}}{2\times 12}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 12\left(-15\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-8±\sqrt{64-48\left(-15\right)}}{2\times 12}$$

Multiply -48 times -15.

$$x=\frac{-8±\sqrt{64+720}}{2\times 12}$$

Add 64 to 720.

$$x=\frac{-8±\sqrt{784}}{2\times 12}$$

Take the square root of 784.

$$x=\frac{-8±28}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-8±28}{24}$$

Now solve the equation x=\frac{-8±28}{24} when ± is plus. Add -8 to 28.

$$x=\frac{20}{24}$$

Reduce the fraction \frac{20}{24} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{6}$$

Now solve the equation x=\frac{-8±28}{24} when ± is minus. Subtract 28 from -8.

$$x=-\frac{36}{24}$$

Reduce the fraction \frac{-36}{24} to lowest terms by extracting and canceling out 12.

$$x=-\frac{3}{2}$$

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 12, the least common multiple of 3,4.

$$4\times 2x-12\left(\frac{5}{4}-x^{2}\right)=0$$

Multiply 4 and 2 to get 8.

$$8x-12\left(\frac{5}{4}-x^{2}\right)=0$$

Use the distributive property to multiply -12 by \frac{5}{4}-x^{2}.

$$8x-15+12x^{2}=0$$

Add 15 to both sides. Anything plus zero gives itself.

$$8x+12x^{2}=15$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$12x^{2}+8x=15$$

Divide both sides by 12.

$$\frac{12x^{2}+8x}{12}=\frac{15}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\frac{8}{12}x=\frac{15}{12}$$

Reduce the fraction \frac{8}{12} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{2}{3}x=\frac{15}{12}$$

Reduce the fraction \frac{15}{12} to lowest terms by extracting and canceling out 3.

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

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{5}{4}+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{5}{4}+\frac{1}{9}$$

Add \frac{5}{4} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{49}{36}$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x+\frac{1}{3}=\frac{7}{6} x+\frac{1}{3}=-\frac{7}{6}$$

Subtract \frac{1}{3} from both sides of the equation.

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