Solve 3x^2-15=4x

$$3 x ^ { 2 } - 15 = 4 x$$

Solve for x

$$x = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$$

$$x=3$$

Steps Using Factoring By Grouping

Subtract 4x from both sides.

$$3x^{2}-15-4x=0$$

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

$$3x^{2}-4x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=-4 ab=3\left(-15\right)=-45$$

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 -45.

$$1,-45 3,-15 5,-9$$

Calculate the sum for each pair.

$$1-45=-44 3-15=-12 5-9=-4$$

The solution is the pair that gives sum -4.

$$a=-9 b=5$$

Rewrite 3x^{2}-4x-15 as \left(3x^{2}-9x\right)+\left(5x-15\right).

$$\left(3x^{2}-9x\right)+\left(5x-15\right)$$

Factor out 3x in the first and 5 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$3x^{2}-15-4x=0$$

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.

$$3x^{2}-4x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -4 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 3\left(-15\right)}}{2\times 3}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 3\left(-15\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-4\right)±\sqrt{16-12\left(-15\right)}}{2\times 3}$$

Multiply -12 times -15.

$$x=\frac{-\left(-4\right)±\sqrt{16+180}}{2\times 3}$$

Add 16 to 180.

$$x=\frac{-\left(-4\right)±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$x=\frac{-\left(-4\right)±14}{2\times 3}$$

The opposite of -4 is 4.

$$x=\frac{4±14}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{4±14}{6}$$

Now solve the equation x=\frac{4±14}{6} when ± is plus. Add 4 to 14.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{4±14}{6} when ± is minus. Subtract 14 from 4.

$$x=-\frac{10}{6}$$

Reduce the fraction \frac{-10}{6} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 4x from both sides.

$$3x^{2}-15-4x=0$$

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

$$3x^{2}-4x=15$$

Divide both sides by 3.

$$\frac{3x^{2}-4x}{3}=\frac{15}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{4}{3}x=\frac{15}{3}$$

Divide 15 by 3.

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

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

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

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

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

Add 5 to \frac{4}{9}.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{49}{9}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{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}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$x-\frac{2}{3}=\frac{7}{3} x-\frac{2}{3}=-\frac{7}{3}$$

Add \frac{2}{3} to both sides of the equation.

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