Solve 3x^2=48-10x

$$3 x ^ { 2 } = 48 - 10 x$$

Solve for x

$$x=-6$$

$$x = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

Steps Using Factoring By Grouping

Subtract 48 from both sides.

$$3x^{2}-48=-10x$$

Add 10x to both sides.

$$3x^{2}-48+10x=0$$

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

$$3x^{2}+10x-48=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-48. To find a and b, set up a system to be solved.

$$a+b=10 ab=3\left(-48\right)=-144$$

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 -144.

$$-1,144 -2,72 -3,48 -4,36 -6,24 -8,18 -9,16 -12,12$$

Calculate the sum for each pair.

$$-1+144=143 -2+72=70 -3+48=45 -4+36=32 -6+24=18 -8+18=10 -9+16=7 -12+12=0$$

The solution is the pair that gives sum 10.

$$a=-8 b=18$$

Rewrite 3x^{2}+10x-48 as \left(3x^{2}-8x\right)+\left(18x-48\right).

$$\left(3x^{2}-8x\right)+\left(18x-48\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(3x-8\right)+6\left(3x-8\right)$$

Factor out common term 3x-8 by using distributive property.

$$\left(3x-8\right)\left(x+6\right)$$

To find equation solutions, solve 3x-8=0 and x+6=0.

$$x=\frac{8}{3} x=-6$$

Steps Using the Quadratic Formula

Subtract 48 from both sides.

$$3x^{2}-48=-10x$$

Add 10x to both sides.

$$3x^{2}-48+10x=0$$

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.

$$3x^{2}+10x-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 10 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 3\left(-48\right)}}{2\times 3}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 3\left(-48\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-10±\sqrt{100-12\left(-48\right)}}{2\times 3}$$

Multiply -12 times -48.

$$x=\frac{-10±\sqrt{100+576}}{2\times 3}$$

Add 100 to 576.

$$x=\frac{-10±\sqrt{676}}{2\times 3}$$

Take the square root of 676.

$$x=\frac{-10±26}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-10±26}{6}$$

Now solve the equation x=\frac{-10±26}{6} when ± is plus. Add -10 to 26.

$$x=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{8}{3}$$

Now solve the equation x=\frac{-10±26}{6} when ± is minus. Subtract 26 from -10.

$$x=-\frac{36}{6}$$

Divide -36 by 6.

$$x=-6$$

The equation is now solved.

$$x=\frac{8}{3} x=-6$$

Steps for Completing the Square

Add 10x to both sides.

$$3x^{2}+10x=48$$

Divide both sides by 3.

$$\frac{3x^{2}+10x}{3}=\frac{48}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{10}{3}x=\frac{48}{3}$$

Divide 48 by 3.

$$x^{2}+\frac{10}{3}x=16$$

Divide \frac{10}{3}, the coefficient of the x term, by 2 to get \frac{5}{3}. Then add the square of \frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=16+\left(\frac{5}{3}\right)^{2}$$

Square \frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=16+\frac{25}{9}$$

Add 16 to \frac{25}{9}.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{169}{9}$$

Factor x^{2}+\frac{10}{3}x+\frac{25}{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}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{169}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{169}{9}}$$

Simplify.

$$x+\frac{5}{3}=\frac{13}{3} x+\frac{5}{3}=-\frac{13}{3}$$

Subtract \frac{5}{3} from both sides of the equation.

$$x=\frac{8}{3} x=-6$$