Solve left(3/4-1right)x^2-2*3/4x+3/4-3=0

$$\left( \frac{ 3 }{ 4 } -1 \right) { x }^{ 2 } - \frac{ 2 \times 3 }{ 4 } x+ \frac{ 3 }{ 4 } -3 = 0$$

Solve for x

$$x=-3$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 4.

$$4\left(\left(\frac{3}{4}-1\right)x^{2}-\frac{2\times 3}{4}x\right)+3-12=0$$

Subtract 1 from \frac{3}{4} to get -\frac{1}{4}.

$$4\left(-\frac{1}{4}x^{2}-\frac{2\times 3}{4}x\right)+3-12=0$$

Multiply 2 and 3 to get 6.

$$4\left(-\frac{1}{4}x^{2}-\frac{6}{4}x\right)+3-12=0$$

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

$$4\left(-\frac{1}{4}x^{2}-\frac{3}{2}x\right)+3-12=0$$

Use the distributive property to multiply 4 by -\frac{1}{4}x^{2}-\frac{3}{2}x.

$$-x^{2}-6x+3-12=0$$

Subtract 12 from 3 to get -9.

$$-x^{2}-6x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=-6 ab=-\left(-9\right)=9$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 9.

$$-1,-9 -3,-3$$

Calculate the sum for each pair.

$$-1-9=-10 -3-3=-6$$

The solution is the pair that gives sum -6.

$$a=-3 b=-3$$

Rewrite -x^{2}-6x-9 as \left(-x^{2}-3x\right)+\left(-3x-9\right).

$$\left(-x^{2}-3x\right)+\left(-3x-9\right)$$

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

$$x\left(-x-3\right)+3\left(-x-3\right)$$

Factor out common term -x-3 by using distributive property.

$$\left(-x-3\right)\left(x+3\right)$$

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

$$x=-3 x=-3$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$4\left(\left(\frac{3}{4}-1\right)x^{2}-\frac{2\times 3}{4}x\right)+3-12=0$$

Subtract 1 from \frac{3}{4} to get -\frac{1}{4}.

$$4\left(-\frac{1}{4}x^{2}-\frac{2\times 3}{4}x\right)+3-12=0$$

Multiply 2 and 3 to get 6.

$$4\left(-\frac{1}{4}x^{2}-\frac{6}{4}x\right)+3-12=0$$

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

$$4\left(-\frac{1}{4}x^{2}-\frac{3}{2}x\right)+3-12=0$$

Use the distributive property to multiply 4 by -\frac{1}{4}x^{2}-\frac{3}{2}x.

$$-x^{2}-6x+3-12=0$$

Subtract 12 from 3 to get -9.

$$-x^{2}-6x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -6 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-1\right)\left(-9\right)}}{2\left(-1\right)}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-1\right)\left(-9\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-6\right)±\sqrt{36+4\left(-9\right)}}{2\left(-1\right)}$$

Multiply 4 times -9.

$$x=\frac{-\left(-6\right)±\sqrt{36-36}}{2\left(-1\right)}$$

Add 36 to -36.

$$x=\frac{-\left(-6\right)±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

$$x=-\frac{-6}{2\left(-1\right)}$$

The opposite of -6 is 6.

$$x=\frac{6}{2\left(-1\right)}$$

Multiply 2 times -1.

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

Divide 6 by -2.

$$x=-3$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$4\left(\left(\frac{3}{4}-1\right)x^{2}-\frac{2\times 3}{4}x\right)+3-12=0$$

Subtract 1 from \frac{3}{4} to get -\frac{1}{4}.

$$4\left(-\frac{1}{4}x^{2}-\frac{2\times 3}{4}x\right)+3-12=0$$

Multiply 2 and 3 to get 6.

$$4\left(-\frac{1}{4}x^{2}-\frac{6}{4}x\right)+3-12=0$$

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

$$4\left(-\frac{1}{4}x^{2}-\frac{3}{2}x\right)+3-12=0$$

Use the distributive property to multiply 4 by -\frac{1}{4}x^{2}-\frac{3}{2}x.

$$-x^{2}-6x+3-12=0$$

Subtract 12 from 3 to get -9.

$$-x^{2}-6x-9=0$$

Add 9 to both sides. Anything plus zero gives itself.

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

Divide both sides by -1.

$$\frac{-x^{2}-6x}{-1}=\frac{9}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{6}{-1}\right)x=\frac{9}{-1}$$

Divide -6 by -1.

$$x^{2}+6x=\frac{9}{-1}$$

Divide 9 by -1.

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

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=-9+3^{2}$$

Square 3.

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

Add -9 to 9.

$$x^{2}+6x+9=0$$

Factor x^{2}+6x+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}.

$$\left(x+3\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+3=0 x+3=0$$

Subtract 3 from both sides of the equation.

$$x=-3 x=-3$$

The equation is now solved. Solutions are the same.

$$x=-3$$