Solve 10x^2-49x=-18

$$10 x ^ { 2 } - 49 x = - 18$$

Solve for x

$$x=\frac{2}{5}=0.4$$

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Factoring By Grouping

Add 18 to both sides.

$$10x^{2}-49x+18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10x^{2}+ax+bx+18. To find a and b, set up a system to be solved.

$$a+b=-49 ab=10\times 18=180$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 180.

$$-1,-180 -2,-90 -3,-60 -4,-45 -5,-36 -6,-30 -9,-20 -10,-18 -12,-15$$

Calculate the sum for each pair.

$$-1-180=-181 -2-90=-92 -3-60=-63 -4-45=-49 -5-36=-41 -6-30=-36 -9-20=-29 -10-18=-28 -12-15=-27$$

The solution is the pair that gives sum -49.

$$a=-45 b=-4$$

Rewrite 10x^{2}-49x+18 as \left(10x^{2}-45x\right)+\left(-4x+18\right).

$$\left(10x^{2}-45x\right)+\left(-4x+18\right)$$

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

$$5x\left(2x-9\right)-2\left(2x-9\right)$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(5x-2\right)$$

To find equation solutions, solve 2x-9=0 and 5x-2=0.

$$x=\frac{9}{2} x=\frac{2}{5}$$

Steps Using the Quadratic Formula

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.

$$10x^{2}-49x=-18$$

Add 18 to both sides of the equation.

$$10x^{2}-49x-\left(-18\right)=-18-\left(-18\right)$$

Subtracting -18 from itself leaves 0.

$$10x^{2}-49x-\left(-18\right)=0$$

Subtract -18 from 0.

$$10x^{2}-49x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -49 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-49\right)±\sqrt{\left(-49\right)^{2}-4\times 10\times 18}}{2\times 10}$$

Square -49.

$$x=\frac{-\left(-49\right)±\sqrt{2401-4\times 10\times 18}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-49\right)±\sqrt{2401-40\times 18}}{2\times 10}$$

Multiply -40 times 18.

$$x=\frac{-\left(-49\right)±\sqrt{2401-720}}{2\times 10}$$

Add 2401 to -720.

$$x=\frac{-\left(-49\right)±\sqrt{1681}}{2\times 10}$$

Take the square root of 1681.

$$x=\frac{-\left(-49\right)±41}{2\times 10}$$

The opposite of -49 is 49.

$$x=\frac{49±41}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{49±41}{20}$$

Now solve the equation x=\frac{49±41}{20} when ± is plus. Add 49 to 41.

$$x=\frac{90}{20}$$

Reduce the fraction \frac{90}{20} to lowest terms by extracting and canceling out 10.

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

Now solve the equation x=\frac{49±41}{20} when ± is minus. Subtract 41 from 49.

$$x=\frac{8}{20}$$

Reduce the fraction \frac{8}{20} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

$$x=\frac{9}{2} x=\frac{2}{5}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$10x^{2}-49x=-18$$

Divide both sides by 10.

$$\frac{10x^{2}-49x}{10}=-\frac{18}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}-\frac{49}{10}x=-\frac{18}{10}$$

Reduce the fraction \frac{-18}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{49}{10}x=-\frac{9}{5}$$

Divide -\frac{49}{10}, the coefficient of the x term, by 2 to get -\frac{49}{20}. Then add the square of -\frac{49}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{49}{10}x+\left(-\frac{49}{20}\right)^{2}=-\frac{9}{5}+\left(-\frac{49}{20}\right)^{2}$$

Square -\frac{49}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{49}{10}x+\frac{2401}{400}=-\frac{9}{5}+\frac{2401}{400}$$

Add -\frac{9}{5} to \frac{2401}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{49}{10}x+\frac{2401}{400}=\frac{1681}{400}$$

Factor x^{2}-\frac{49}{10}x+\frac{2401}{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}.

$$\left(x-\frac{49}{20}\right)^{2}=\frac{1681}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{49}{20}\right)^{2}}=\sqrt{\frac{1681}{400}}$$

Simplify.

$$x-\frac{49}{20}=\frac{41}{20} x-\frac{49}{20}=-\frac{41}{20}$$

Add \frac{49}{20} to both sides of the equation.

$$x=\frac{9}{2} x=\frac{2}{5}$$