Answer:
One example is the solution to the equation x2 + 2x − 15 = 0. Substitute values in the formula:
x = [−b ± √(b2 − 4ac)]/2a
a = 1, b = 2, and c = −15. Thus:
x = [−2 ± √(22 − 4*1*{−15})]/2
x = [−2 ± √(4 + 60)]/2
x = [−2 ± √(64)]/2
x = [−2 ± 8]/2
The two solutions are:
x = −10/2 and x = +6/2
x = −5 and x = 3
Example 2
Try the equation 2x2 − x − 1 = 0:
x = [1 ± √(12 − 4*2*{−1})]/4
x = [1 ± √(1 + 8)]/4
Step-by-step explanation:
pa brainslest answer thanks
There's no given quadratic equation. The only given here is the formula for quadratic formula.
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Answer:
One example is the solution to the equation x2 + 2x − 15 = 0. Substitute values in the formula:
x = [−b ± √(b2 − 4ac)]/2a
a = 1, b = 2, and c = −15. Thus:
x = [−2 ± √(22 − 4*1*{−15})]/2
x = [−2 ± √(4 + 60)]/2
x = [−2 ± √(64)]/2
x = [−2 ± 8]/2
The two solutions are:
x = −10/2 and x = +6/2
x = −5 and x = 3
Example 2
Try the equation 2x2 − x − 1 = 0:
x = [−b ± √(b2 − 4ac)]/2a
x = [1 ± √(12 − 4*2*{−1})]/4
x = [1 ± √(1 + 8)]/4
Step-by-step explanation:
pa brainslest answer thanks
Answer:
There's no given quadratic equation. The only given here is the formula for quadratic formula.