Solve -2x^2+35x=143

$$- 2 x ^ { 2 } + 35 x = 143$$

Solve for x

$$x=11$$

$$x = \frac{13}{2} = 6\frac{1}{2} = 6.5$$

Steps Using Factoring By Grouping

Subtract 143 from both sides.

$$-2x^{2}+35x-143=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-143. To find a and b, set up a system to be solved.

$$a+b=35 ab=-2\left(-143\right)=286$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 286.

$$1,286 2,143 11,26 13,22$$

Calculate the sum for each pair.

$$1+286=287 2+143=145 11+26=37 13+22=35$$

The solution is the pair that gives sum 35.

$$a=22 b=13$$

Rewrite -2x^{2}+35x-143 as \left(-2x^{2}+22x\right)+\left(13x-143\right).

$$\left(-2x^{2}+22x\right)+\left(13x-143\right)$$

Factor out 2x in the first and -13 in the second group.

$$2x\left(-x+11\right)-13\left(-x+11\right)$$

Factor out common term -x+11 by using distributive property.

$$\left(-x+11\right)\left(2x-13\right)$$

To find equation solutions, solve -x+11=0 and 2x-13=0.

$$x=11 x=\frac{13}{2}$$

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.

$$-2x^{2}+35x=143$$

Subtract 143 from both sides of the equation.

$$-2x^{2}+35x-143=143-143$$

Subtracting 143 from itself leaves 0.

$$-2x^{2}+35x-143=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 35 for b, and -143 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-35±\sqrt{35^{2}-4\left(-2\right)\left(-143\right)}}{2\left(-2\right)}$$

Square 35.

$$x=\frac{-35±\sqrt{1225-4\left(-2\right)\left(-143\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-35±\sqrt{1225+8\left(-143\right)}}{2\left(-2\right)}$$

Multiply 8 times -143.

$$x=\frac{-35±\sqrt{1225-1144}}{2\left(-2\right)}$$

Add 1225 to -1144.

$$x=\frac{-35±\sqrt{81}}{2\left(-2\right)}$$

Take the square root of 81.

$$x=\frac{-35±9}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-35±9}{-4}$$

Now solve the equation x=\frac{-35±9}{-4} when ± is plus. Add -35 to 9.

$$x=-\frac{26}{-4}$$

Reduce the fraction \frac{-26}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{13}{2}$$

Now solve the equation x=\frac{-35±9}{-4} when ± is minus. Subtract 9 from -35.

$$x=-\frac{44}{-4}$$

Divide -44 by -4.

$$x=11$$

The equation is now solved.

$$x=\frac{13}{2} x=11$$

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.

$$-2x^{2}+35x=143$$

Divide both sides by -2.

$$\frac{-2x^{2}+35x}{-2}=\frac{143}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{35}{-2}x=\frac{143}{-2}$$

Divide 35 by -2.

$$x^{2}-\frac{35}{2}x=\frac{143}{-2}$$

Divide 143 by -2.

$$x^{2}-\frac{35}{2}x=-\frac{143}{2}$$

Divide -\frac{35}{2}, the coefficient of the x term, by 2 to get -\frac{35}{4}. Then add the square of -\frac{35}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{35}{2}x+\left(-\frac{35}{4}\right)^{2}=-\frac{143}{2}+\left(-\frac{35}{4}\right)^{2}$$

Square -\frac{35}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{35}{2}x+\frac{1225}{16}=-\frac{143}{2}+\frac{1225}{16}$$

Add -\frac{143}{2} to \frac{1225}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{35}{2}x+\frac{1225}{16}=\frac{81}{16}$$

Factor x^{2}-\frac{35}{2}x+\frac{1225}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{35}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{35}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x-\frac{35}{4}=\frac{9}{4} x-\frac{35}{4}=-\frac{9}{4}$$

Add \frac{35}{4} to both sides of the equation.

$$x=11 x=\frac{13}{2}$$