Solve 3x(2x+1)-5=-10x

$$3 x ( 2 x + 1 ) - 5 = - 10 x$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x=\frac{1}{3}\approx 0.333333333$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 3x by 2x+1.

$$6x^{2}+3x-5=-10x$$

Add 10x to both sides.

$$6x^{2}+3x-5+10x=0$$

Combine 3x and 10x to get 13x.

$$6x^{2}+13x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=13 ab=6\left(-5\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 13.

$$a=-2 b=15$$

Rewrite 6x^{2}+13x-5 as \left(6x^{2}-2x\right)+\left(15x-5\right).

$$\left(6x^{2}-2x\right)+\left(15x-5\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(3x-1\right)+5\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(2x+5\right)$$

To find equation solutions, solve 3x-1=0 and 2x+5=0.

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

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by 2x+1.

$$6x^{2}+3x-5=-10x$$

Add 10x to both sides.

$$6x^{2}+3x-5+10x=0$$

Combine 3x and 10x to get 13x.

$$6x^{2}+13x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 13 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}-4\times 6\left(-5\right)}}{2\times 6}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\times 6\left(-5\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-13±\sqrt{169-24\left(-5\right)}}{2\times 6}$$

Multiply -24 times -5.

$$x=\frac{-13±\sqrt{169+120}}{2\times 6}$$

Add 169 to 120.

$$x=\frac{-13±\sqrt{289}}{2\times 6}$$

Take the square root of 289.

$$x=\frac{-13±17}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-13±17}{12}$$

Now solve the equation x=\frac{-13±17}{12} when ± is plus. Add -13 to 17.

$$x=\frac{4}{12}$$

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

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

Now solve the equation x=\frac{-13±17}{12} when ± is minus. Subtract 17 from -13.

$$x=-\frac{30}{12}$$

Reduce the fraction \frac{-30}{12} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply 3x by 2x+1.

$$6x^{2}+3x-5=-10x$$

Add 10x to both sides.

$$6x^{2}+3x-5+10x=0$$

Combine 3x and 10x to get 13x.

$$6x^{2}+13x-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$6x^{2}+13x=5$$

Divide both sides by 6.

$$\frac{6x^{2}+13x}{6}=\frac{5}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\frac{13}{6}x=\frac{5}{6}$$

Divide \frac{13}{6}, the coefficient of the x term, by 2 to get \frac{13}{12}. Then add the square of \frac{13}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{13}{6}x+\left(\frac{13}{12}\right)^{2}=\frac{5}{6}+\left(\frac{13}{12}\right)^{2}$$

Square \frac{13}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{13}{6}x+\frac{169}{144}=\frac{5}{6}+\frac{169}{144}$$

Add \frac{5}{6} to \frac{169}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{13}{6}x+\frac{169}{144}=\frac{289}{144}$$

Factor x^{2}+\frac{13}{6}x+\frac{169}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{13}{12}\right)^{2}=\frac{289}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{12}\right)^{2}}=\sqrt{\frac{289}{144}}$$

Simplify.

$$x+\frac{13}{12}=\frac{17}{12} x+\frac{13}{12}=-\frac{17}{12}$$

Subtract \frac{13}{12} from both sides of the equation.

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