Solve x(x + 5) = 15
Quadratic equation, worked out line by line the way a teacher would write it.
Answer
| Solutions | x = (−√85 − 5)/2, x = (√85 − 5)/2 |
Step-by-step solution
10 steps-
1 Givenx \left(x + 5\right) = 15
Solve for x.
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2 Distribute: multiply each term inside the brackets
The distributive property, a(b + c) = ab + ac: the number in front of a bracket multiplies every term inside it, not just the first. A minus in front works like −1, so it flips the sign of every term inside.
x^{2} + 5 x = 15 -
3 Move every term to the left side so the right side is 0
Factoring and the quadratic formula both work on an equation of the form … = 0. Subtracting the right side from both sides gets there without changing the solutions.
x^{2} + 5 x - 15 = 0 -
4 This is a quadratic in standard form ax² + bx + c = 0
Every quadratic equation can be arranged as ax² + bx + c = 0. Reading off a = 1, b = 5 and c = -15, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = 5,\quad c = -15 -
5 Apply the quadratic formula
The quadratic formula solves any equation ax² + bx + c = 0. It comes from completing the square on the general equation, so it always works, even when the trinomial does not factor.
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} -
6 Substitute a, b and cx = \frac{-5 \pm \sqrt{5^{2} - 4 \cdot 1 \cdot (-15)}}{2 \cdot 1}
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7 Compute the discriminant D = b² − 4ac
D = b² − 4ac is the part under the square root in the formula. If it is positive there are two real solutions, if it is 0 there is one, and if it is negative its square root is not a real number, so there is no real solution.
D = 85Positive discriminant: two distinct real solutions.
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8 Evaluate the square root and the denominatorx = \frac{-5 \pm \sqrt{85}}{2}
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9 Simplifyx = \frac{- \sqrt{85} - 5}{2}\quad \text{or} \quad x = \frac{\sqrt{85} - 5}{2}
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10 Check\begin{aligned}x = \frac{- \sqrt{85} - 5}{2}:\quad 15 = 15\quad\checkmark\\ x = \frac{\sqrt{85} - 5}{2}:\quad 15 = 15\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| (−√85 − 5)/2 | 15 | 15 | ✓ |
| (√85 − 5)/2 | 15 | 15 | ✓ |
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