Question 380277: how can i solve these two problems with the quadratic formula.
6x^2+x-2=0
3x^2+18x+15=0
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! "how can i solve these two problems with the quadratic formula.
6x^2+x-2=0
3x^2+18x+15=0"
quadratic formula:
quadratic equation of form ax^2 + bx + c = 0
first quadratic equation: 6x^2 + x - 2 = 0
a = 6, b = 1, c = -2
2nd quadratic equation: 3x^2 + 18x + 15 = 0
a = 3, b = 18, c = 15
plug in values and solve
discriminant first:
b^2 - 4ac = 1 - 4 * 6 * -2 = 1 - 24 * -2 = 1 - -48
b^2 - 4ac = 1 + 48 = 49, so 2 real roots
discriminant second:
b^2 - 4ac = 18^2 - 4 * 3 * 15 = 324 - 12 * 15
b^2 - 4ac = 324 - 180 = 144, so 2 real roots
first:
x1 = -1/12 + 7/12 = 6/12 = 1/2
x2 = -1/12 - 7/12 = -8/12 = -2/3
check: 6x^2 + x - 2 = 0
6(1/2)^2 + (1/2) - 2 = 6 * 1/4 + 1/2 - 2 = 3/2 + 1/2 - 2
6(1/2)^2 + (1/2) - 2 = 4/2 - 2 = 2 - 2 = 0, yes
6(-2/3)^2 + (-2/3) - 2 = 6 * 4/9 - 2/3 - 2 = 24/9 - 6/9 - 2
6(-2/3)^2 + (-2/3) - 2 = 18/9 - 2 = 2 - 2 = 0, yes
second:
x1 = -18/6 + 12/6 = -6/6 = -1
x2 = -18/6 - 12/6 = -30/6 = -5
check: 3x^2 + 18x + 15 = 0
3*-1^2 - 18 + 15 = 3 * 1 - 3 = 3 - 3 = 0, yes
3*-5^2 + 18(-5) + 15 = 3 * 25 - 90 + 15 = 75 - 75 = 0, yes
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