Solve -2x^2+47x+5=275

$$- 2 x ^ { 2 } + 47 x + 5 = 275$$

Solve for x

$$x=10$$

$$x = \frac{27}{2} = 13\frac{1}{2} = 13.5$$

Steps Using Factoring By Grouping

Subtract 275 from both sides.

$$-2x^{2}+47x+5-275=0$$

Subtract 275 from 5 to get -270.

$$-2x^{2}+47x-270=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-270. To find a and b, set up a system to be solved.

$$a+b=47 ab=-2\left(-270\right)=540$$

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 540.

$$1,540 2,270 3,180 4,135 5,108 6,90 9,60 10,54 12,45 15,36 18,30 20,27$$

Calculate the sum for each pair.

$$1+540=541 2+270=272 3+180=183 4+135=139 5+108=113 6+90=96 9+60=69 10+54=64 12+45=57 15+36=51 18+30=48 20+27=47$$

The solution is the pair that gives sum 47.

$$a=27 b=20$$

Rewrite -2x^{2}+47x-270 as \left(-2x^{2}+27x\right)+\left(20x-270\right).

$$\left(-2x^{2}+27x\right)+\left(20x-270\right)$$

Factor out -x in the first and 10 in the second group.

$$-x\left(2x-27\right)+10\left(2x-27\right)$$

Factor out common term 2x-27 by using distributive property.

$$\left(2x-27\right)\left(-x+10\right)$$

To find equation solutions, solve 2x-27=0 and -x+10=0.

$$x=\frac{27}{2} x=10$$

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.

$$-2x^{2}+47x+5=275$$

Subtract 275 from both sides of the equation.

$$-2x^{2}+47x+5-275=275-275$$

Subtracting 275 from itself leaves 0.

$$-2x^{2}+47x+5-275=0$$

Subtract 275 from 5.

$$-2x^{2}+47x-270=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 47 for b, and -270 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-47±\sqrt{47^{2}-4\left(-2\right)\left(-270\right)}}{2\left(-2\right)}$$

Square 47.

$$x=\frac{-47±\sqrt{2209-4\left(-2\right)\left(-270\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-47±\sqrt{2209+8\left(-270\right)}}{2\left(-2\right)}$$

Multiply 8 times -270.

$$x=\frac{-47±\sqrt{2209-2160}}{2\left(-2\right)}$$

Add 2209 to -2160.

$$x=\frac{-47±\sqrt{49}}{2\left(-2\right)}$$

Take the square root of 49.

$$x=\frac{-47±7}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-47±7}{-4}$$

Now solve the equation x=\frac{-47±7}{-4} when ± is plus. Add -47 to 7.

$$x=-\frac{40}{-4}$$

Divide -40 by -4.

$$x=10$$

Now solve the equation x=\frac{-47±7}{-4} when ± is minus. Subtract 7 from -47.

$$x=-\frac{54}{-4}$$

Reduce the fraction \frac{-54}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{27}{2}$$

The equation is now solved.

$$x=10 x=\frac{27}{2}$$

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.

$$-2x^{2}+47x+5=275$$

Subtract 5 from both sides of the equation.

$$-2x^{2}+47x+5-5=275-5$$

Subtracting 5 from itself leaves 0.

$$-2x^{2}+47x=275-5$$

Subtract 5 from 275.

$$-2x^{2}+47x=270$$

Divide both sides by -2.

$$\frac{-2x^{2}+47x}{-2}=\frac{270}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{47}{-2}x=\frac{270}{-2}$$

Divide 47 by -2.

$$x^{2}-\frac{47}{2}x=\frac{270}{-2}$$

Divide 270 by -2.

$$x^{2}-\frac{47}{2}x=-135$$

Divide -\frac{47}{2}, the coefficient of the x term, by 2 to get -\frac{47}{4}. Then add the square of -\frac{47}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{47}{2}x+\left(-\frac{47}{4}\right)^{2}=-135+\left(-\frac{47}{4}\right)^{2}$$

Square -\frac{47}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{47}{2}x+\frac{2209}{16}=-135+\frac{2209}{16}$$

Add -135 to \frac{2209}{16}.

$$x^{2}-\frac{47}{2}x+\frac{2209}{16}=\frac{49}{16}$$

Factor x^{2}-\frac{47}{2}x+\frac{2209}{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}.

$$\left(x-\frac{47}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{47}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x-\frac{47}{4}=\frac{7}{4} x-\frac{47}{4}=-\frac{7}{4}$$

Add \frac{47}{4} to both sides of the equation.

$$x=\frac{27}{2} x=10$$