Solve 10x^2+3x-27=2x-3

$$10 x ^ { 2 } + 3 x - 27 = 2 x - 3$$

Solve for x

$$x = -\frac{8}{5} = -1\frac{3}{5} = -1.6$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Subtract 2x from both sides.

$$10x^{2}+3x-27-2x=-3$$

Combine 3x and -2x to get x.

$$10x^{2}+x-27=-3$$

Add 3 to both sides.

$$10x^{2}+x-27+3=0$$

Add -27 and 3 to get -24.

$$10x^{2}+x-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=1 ab=10\left(-24\right)=-240$$

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 -240.

$$-1,240 -2,120 -3,80 -4,60 -5,48 -6,40 -8,30 -10,24 -12,20 -15,16$$

Calculate the sum for each pair.

$$-1+240=239 -2+120=118 -3+80=77 -4+60=56 -5+48=43 -6+40=34 -8+30=22 -10+24=14 -12+20=8 -15+16=1$$

The solution is the pair that gives sum 1.

$$a=-15 b=16$$

Rewrite 10x^{2}+x-24 as \left(10x^{2}-15x\right)+\left(16x-24\right).

$$\left(10x^{2}-15x\right)+\left(16x-24\right)$$

Factor out 5x in the first and 8 in the second group.

$$5x\left(2x-3\right)+8\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(5x+8\right)$$

To find equation solutions, solve 2x-3=0 and 5x+8=0.

$$x=\frac{3}{2} x=-\frac{8}{5}$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

$$10x^{2}+3x-27-2x=-3$$

Combine 3x and -2x to get x.

$$10x^{2}+x-27=-3$$

Add 3 to both sides.

$$10x^{2}+x-27+3=0$$

Add -27 and 3 to get -24.

$$10x^{2}+x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 1 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 10\left(-24\right)}}{2\times 10}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 10\left(-24\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-1±\sqrt{1-40\left(-24\right)}}{2\times 10}$$

Multiply -40 times -24.

$$x=\frac{-1±\sqrt{1+960}}{2\times 10}$$

Add 1 to 960.

$$x=\frac{-1±\sqrt{961}}{2\times 10}$$

Take the square root of 961.

$$x=\frac{-1±31}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-1±31}{20}$$

Now solve the equation x=\frac{-1±31}{20} when ± is plus. Add -1 to 31.

$$x=\frac{30}{20}$$

Reduce the fraction \frac{30}{20} to lowest terms by extracting and canceling out 10.

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

Now solve the equation x=\frac{-1±31}{20} when ± is minus. Subtract 31 from -1.

$$x=-\frac{32}{20}$$

Reduce the fraction \frac{-32}{20} to lowest terms by extracting and canceling out 4.

$$x=-\frac{8}{5}$$

The equation is now solved.

$$x=\frac{3}{2} x=-\frac{8}{5}$$

Steps for Completing the Square

Subtract 2x from both sides.

$$10x^{2}+3x-27-2x=-3$$

Combine 3x and -2x to get x.

$$10x^{2}+x-27=-3$$

Add 27 to both sides.

$$10x^{2}+x=-3+27$$

Add -3 and 27 to get 24.

$$10x^{2}+x=24$$

Divide both sides by 10.

$$\frac{10x^{2}+x}{10}=\frac{24}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\frac{1}{10}x=\frac{24}{10}$$

Reduce the fraction \frac{24}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{10}x=\frac{12}{5}$$

Divide \frac{1}{10}, the coefficient of the x term, by 2 to get \frac{1}{20}. Then add the square of \frac{1}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{10}x+\left(\frac{1}{20}\right)^{2}=\frac{12}{5}+\left(\frac{1}{20}\right)^{2}$$

Square \frac{1}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{10}x+\frac{1}{400}=\frac{12}{5}+\frac{1}{400}$$

Add \frac{12}{5} to \frac{1}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{10}x+\frac{1}{400}=\frac{961}{400}$$

Factor x^{2}+\frac{1}{10}x+\frac{1}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{20}\right)^{2}=\frac{961}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{20}\right)^{2}}=\sqrt{\frac{961}{400}}$$

Simplify.

$$x+\frac{1}{20}=\frac{31}{20} x+\frac{1}{20}=-\frac{31}{20}$$

Subtract \frac{1}{20} from both sides of the equation.

$$x=\frac{3}{2} x=-\frac{8}{5}$$