Solve 3x-3=-4x^2+14x

$$3x-3=-4 { x }^{ 2 } +14x$$

Solve for x

$$x=-\frac{1}{4}=-0.25$$

$$x=3$$

Steps Using Factoring By Grouping

Add 4x^{2} to both sides.

$$3x-3+4x^{2}=14x$$

Subtract 14x from both sides.

$$3x-3+4x^{2}-14x=0$$

Combine 3x and -14x to get -11x.

$$-11x-3+4x^{2}=0$$

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

$$4x^{2}-11x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-11 ab=4\left(-3\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -11.

$$a=-12 b=1$$

Rewrite 4x^{2}-11x-3 as \left(4x^{2}-12x\right)+\left(x-3\right).

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

Factor out 4x in 4x^{2}-12x.

$$4x\left(x-3\right)+x-3$$

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

$$\left(x-3\right)\left(4x+1\right)$$

To find equation solutions, solve x-3=0 and 4x+1=0.

$$x=3 x=-\frac{1}{4}$$

Steps Using the Quadratic Formula

Add 4x^{2} to both sides.

$$3x-3+4x^{2}=14x$$

Subtract 14x from both sides.

$$3x-3+4x^{2}-14x=0$$

Combine 3x and -14x to get -11x.

$$-11x-3+4x^{2}=0$$

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.

$$4x^{2}-11x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -11 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 4\left(-3\right)}}{2\times 4}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 4\left(-3\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-11\right)±\sqrt{121-16\left(-3\right)}}{2\times 4}$$

Multiply -16 times -3.

$$x=\frac{-\left(-11\right)±\sqrt{121+48}}{2\times 4}$$

Add 121 to 48.

$$x=\frac{-\left(-11\right)±\sqrt{169}}{2\times 4}$$

Take the square root of 169.

$$x=\frac{-\left(-11\right)±13}{2\times 4}$$

The opposite of -11 is 11.

$$x=\frac{11±13}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{11±13}{8}$$

Now solve the equation x=\frac{11±13}{8} when ± is plus. Add 11 to 13.

$$x=\frac{24}{8}$$

Divide 24 by 8.

$$x=3$$

Now solve the equation x=\frac{11±13}{8} when ± is minus. Subtract 13 from 11.

$$x=-\frac{2}{8}$$

Reduce the fraction \frac{-2}{8} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{4}$$

The equation is now solved.

$$x=3 x=-\frac{1}{4}$$

Steps for Completing the Square

Add 4x^{2} to both sides.

$$3x-3+4x^{2}=14x$$

Subtract 14x from both sides.

$$3x-3+4x^{2}-14x=0$$

Combine 3x and -14x to get -11x.

$$-11x-3+4x^{2}=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$-11x+4x^{2}=3$$

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.

$$4x^{2}-11x=3$$

Divide both sides by 4.

$$\frac{4x^{2}-11x}{4}=\frac{3}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{11}{4}x=\frac{3}{4}$$

Divide -\frac{11}{4}, the coefficient of the x term, by 2 to get -\frac{11}{8}. Then add the square of -\frac{11}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{4}x+\left(-\frac{11}{8}\right)^{2}=\frac{3}{4}+\left(-\frac{11}{8}\right)^{2}$$

Square -\frac{11}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{4}x+\frac{121}{64}=\frac{3}{4}+\frac{121}{64}$$

Add \frac{3}{4} to \frac{121}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{4}x+\frac{121}{64}=\frac{169}{64}$$

Factor x^{2}-\frac{11}{4}x+\frac{121}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{8}\right)^{2}=\frac{169}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{8}\right)^{2}}=\sqrt{\frac{169}{64}}$$

Simplify.

$$x-\frac{11}{8}=\frac{13}{8} x-\frac{11}{8}=-\frac{13}{8}$$

Add \frac{11}{8} to both sides of the equation.

$$x=3 x=-\frac{1}{4}$$