Solve -x² + 5x + 7 = 0
The solutions of -x² + 5x + 7 = 0 are x = (5 + √53)/2 or x = (5 - √53)/2, that is x ≈ -1.1400549 and x ≈ 6.1400549.
Change a, b or c (whole numbers, a not 0) and the exact solutions follow.
How to work it out
- 1
Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = -1, b = 5, c = 7.
- 2
Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.
- 3
D is positive, so there are two different real solutions.
- 4
The quadratic formula works for every quadratic:
- 5
Simplify (pull the square out of the root) to get the solutions:
- 6
As decimals: x ≈ -1.1400549 and x ≈ 6.1400549.
- 7
Check numerically: for x ≈ -1.1400549 the left side is 1.7763568e-15 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ -x^2 + 5x + 7 = 0\ \Rightarrow\ x = \frac{5 + \sqrt{53}}{2},\ \frac{5 - \sqrt{53}}{2} \]
\[ -x^2 + 5x + 7 = 0 \]
\[ D = 5^2 - 4\cdot(-1)\cdot7 = 25 + 28 = 53 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{53}}{-2} \]
\[ x = \frac{5 + \sqrt{53}}{2}\quad\text{or}\quad x = \frac{5 - \sqrt{53}}{2} \]The parabola y = -x² + 5x + 7
| x | y |
|---|---|
| -0.5 | 4.25 |
| 0.5 | 9.25 |
| 1.5 | 12.25 |
| 2.5 | 13.25 |
| 3.5 | 12.25 |
| 4.5 | 9.25 |
| 5.5 | 4.25 |
Questions
What are the solutions of -x² + 5x + 7 = 0?
The solutions of -x² + 5x + 7 = 0 are x = (5 + √53)/2 or x = (5 - √53)/2, that is x ≈ -1.1400549 and x ≈ 6.1400549.
Does -x² + 5x + 7 = 0 factor?
Not over the integers, because the discriminant 53 is not a perfect square. Use the quadratic formula.
What is the vertex of y = -x² + 5x + 7?
(5/2, 53/4). The parabola opens downwards, so this is the maximum.
What are the sum and product of the roots?
Sum = −b/a = 5, product = c/a = -7.