Solve 10x^2=-17x+20

$$10 x ^ { 2 } = - 17 x + 20$$

Solve for x

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

$$x=\frac{4}{5}=0.8$$

Steps Using Factoring By Grouping

Add 17x to both sides.

$$10x^{2}+17x=20$$

Subtract 20 from both sides.

$$10x^{2}+17x-20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=17 ab=10\left(-20\right)=-200$$

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 -200.

$$-1,200 -2,100 -4,50 -5,40 -8,25 -10,20$$

Calculate the sum for each pair.

$$-1+200=199 -2+100=98 -4+50=46 -5+40=35 -8+25=17 -10+20=10$$

The solution is the pair that gives sum 17.

$$a=-8 b=25$$

Rewrite 10x^{2}+17x-20 as \left(10x^{2}-8x\right)+\left(25x-20\right).

$$\left(10x^{2}-8x\right)+\left(25x-20\right)$$

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

$$2x\left(5x-4\right)+5\left(5x-4\right)$$

Factor out common term 5x-4 by using distributive property.

$$\left(5x-4\right)\left(2x+5\right)$$

To find equation solutions, solve 5x-4=0 and 2x+5=0.

$$x=\frac{4}{5} x=-\frac{5}{2}$$

Steps Using the Quadratic Formula

Add 17x to both sides.

$$10x^{2}+17x=20$$

Subtract 20 from both sides.

$$10x^{2}+17x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 17 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-17±\sqrt{17^{2}-4\times 10\left(-20\right)}}{2\times 10}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\times 10\left(-20\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-17±\sqrt{289-40\left(-20\right)}}{2\times 10}$$

Multiply -40 times -20.

$$x=\frac{-17±\sqrt{289+800}}{2\times 10}$$

Add 289 to 800.

$$x=\frac{-17±\sqrt{1089}}{2\times 10}$$

Take the square root of 1089.

$$x=\frac{-17±33}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-17±33}{20}$$

Now solve the equation x=\frac{-17±33}{20} when ± is plus. Add -17 to 33.

$$x=\frac{16}{20}$$

Reduce the fraction \frac{16}{20} to lowest terms by extracting and canceling out 4.

$$x=\frac{4}{5}$$

Now solve the equation x=\frac{-17±33}{20} when ± is minus. Subtract 33 from -17.

$$x=-\frac{50}{20}$$

Reduce the fraction \frac{-50}{20} to lowest terms by extracting and canceling out 10.

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

The equation is now solved.

$$x=\frac{4}{5} x=-\frac{5}{2}$$

Steps for Completing the Square

Add 17x to both sides.

$$10x^{2}+17x=20$$

Divide both sides by 10.

$$\frac{10x^{2}+17x}{10}=\frac{20}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\frac{17}{10}x=\frac{20}{10}$$

Divide 20 by 10.

$$x^{2}+\frac{17}{10}x=2$$

Divide \frac{17}{10}, the coefficient of the x term, by 2 to get \frac{17}{20}. Then add the square of \frac{17}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{17}{10}x+\left(\frac{17}{20}\right)^{2}=2+\left(\frac{17}{20}\right)^{2}$$

Square \frac{17}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{17}{10}x+\frac{289}{400}=2+\frac{289}{400}$$

Add 2 to \frac{289}{400}.

$$x^{2}+\frac{17}{10}x+\frac{289}{400}=\frac{1089}{400}$$

Factor x^{2}+\frac{17}{10}x+\frac{289}{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}.

$$\left(x+\frac{17}{20}\right)^{2}=\frac{1089}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{17}{20}\right)^{2}}=\sqrt{\frac{1089}{400}}$$

Simplify.

$$x+\frac{17}{20}=\frac{33}{20} x+\frac{17}{20}=-\frac{33}{20}$$

Subtract \frac{17}{20} from both sides of the equation.

$$x=\frac{4}{5} x=-\frac{5}{2}$$