Answer
(Solution 1, Solution 2) = (27, -28)
↓Steps Explained:↓
Solve the quadratic:
2n2+2n-1512 = 0
Set up the a, b, and c values:
a = 2, b = 2, c = -1512
Quadratic Formula
| n = | -b ± √b2 - 4ac |
| 2a |
Calculate -b
-b = -(2)
-b = -2
Calculate the discriminant Δ
Δ = b2 - 4ac:
Δ = 22 - 4 x 2 x -1512
Δ = 4 - -12096
Δ = 12100 <--- Discriminant
Since Δ > 0, we expect two real roots.
Take the square root of Δ
√Δ = √(12100)
√Δ = 110
-b + Δ:
Numerator 1 = -b + √Δ
Numerator 1 = -2 + 110
Numerator 1 = 108
-b - Δ:
Numerator 2 = -b - √Δ
Numerator 2 = -2 - 110
Numerator 2 = -112
Calculate 2a
Denominator = 2 * a
Denominator = 2 * 2
Denominator = 4
Find Solutions
| Solution 1 = | Numerator 1 |
| Denominator |
Solution 1 = 27
Solution 2
| Solution 2 = | Numerator 2 |
| Denominator |
Solution 2 = -28
Solution Set
(Solution 1, Solution 2) = (27, -28)
Prove our first answer
(27)2 + 2(27) - 1512 ? 0
(729) + 541512 ? 0
1458 + 541512 ? 0
0 = 0
Prove our second answer
(-28)2 + 2(-28) - 1512 ? 0
(784) - 561512 ? 0
1568 - 561512 ? 0
0 = 0
Final Answer
(Solution 1, Solution 2) = (27, -28)Related Calculators: Quartic Equations | 3 Point Equation | Monomials