Solve 3x^2-5x=2x^2+4x+10

$$3 x ^ { 2 } - 5 x = 2 x ^ { 2 } + 4 x + 10$$

Solve for x

$$x=-1$$

$$x=10$$

Steps Using Factoring

Subtract 2x^{2} from both sides.

$$3x^{2}-5x-2x^{2}=4x+10$$

Combine 3x^{2} and -2x^{2} to get x^{2}.

$$x^{2}-5x=4x+10$$

Subtract 4x from both sides.

$$x^{2}-5x-4x=10$$

Combine -5x and -4x to get -9x.

$$x^{2}-9x=10$$

Subtract 10 from both sides.

$$x^{2}-9x-10=0$$

To solve the equation, factor x^{2}-9x-10 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-9 ab=-10$$

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 -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -9.

$$a=-10 b=1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-10\right)\left(x+1\right)$$

To find equation solutions, solve x-10=0 and x+1=0.

$$x=10 x=-1$$

Steps Using Factoring By Grouping

Subtract 2x^{2} from both sides.

$$3x^{2}-5x-2x^{2}=4x+10$$

Combine 3x^{2} and -2x^{2} to get x^{2}.

$$x^{2}-5x=4x+10$$

Subtract 4x from both sides.

$$x^{2}-5x-4x=10$$

Combine -5x and -4x to get -9x.

$$x^{2}-9x=10$$

Subtract 10 from both sides.

$$x^{2}-9x-10=0$$

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

$$a+b=-9 ab=1\left(-10\right)=-10$$

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 -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -9.

$$a=-10 b=1$$

Rewrite x^{2}-9x-10 as \left(x^{2}-10x\right)+\left(x-10\right).

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

Factor out x in x^{2}-10x.

$$x\left(x-10\right)+x-10$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x+1\right)$$

To find equation solutions, solve x-10=0 and x+1=0.

$$x=10 x=-1$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

$$3x^{2}-5x-2x^{2}=4x+10$$

Combine 3x^{2} and -2x^{2} to get x^{2}.

$$x^{2}-5x=4x+10$$

Subtract 4x from both sides.

$$x^{2}-5x-4x=10$$

Combine -5x and -4x to get -9x.

$$x^{2}-9x=10$$

Subtract 10 from both sides.

$$x^{2}-9x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -9 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-10\right)}}{2}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\left(-10\right)}}{2}$$

Multiply -4 times -10.

$$x=\frac{-\left(-9\right)±\sqrt{81+40}}{2}$$

Add 81 to 40.

$$x=\frac{-\left(-9\right)±\sqrt{121}}{2}$$

Take the square root of 121.

$$x=\frac{-\left(-9\right)±11}{2}$$

The opposite of -9 is 9.

$$x=\frac{9±11}{2}$$

Now solve the equation x=\frac{9±11}{2} when ± is plus. Add 9 to 11.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{9±11}{2} when ± is minus. Subtract 11 from 9.

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

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=10 x=-1$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

$$3x^{2}-5x-2x^{2}=4x+10$$

Combine 3x^{2} and -2x^{2} to get x^{2}.

$$x^{2}-5x=4x+10$$

Subtract 4x from both sides.

$$x^{2}-5x-4x=10$$

Combine -5x and -4x to get -9x.

$$x^{2}-9x=10$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=10+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=10+\frac{81}{4}$$

Add 10 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{121}{4}$$

Factor x^{2}-9x+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{11}{2} x-\frac{9}{2}=-\frac{11}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=10 x=-1$$