help on algebra??!?
solve using the quadratic formula, x^2 + 4x - 21 = 0, explaining all steps.
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answers:
Krishnamurthy say: x^2 + 4x - 21 = 0
x^2 + 7x - 3x - 21 = 0
x(x + 7) - 3(x + 7) = 0
(x + 7)(x - 3) = 0
Solutions:
x = -7
x = 3
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Engr. Ronald say: x^2 + 4x - 21 = 0


........- b± √[b^2 - 4ac]
x = -----------------------------
...................2a


.......- 4± √[4^2 - 4(1)(-21)]
x=-----------------------------------
......................2(1)

........ - 4 ± √(100)
x = ----------------------
.............2

......... - 4 ± 10
x = -----------------
..............2

......... - 4 + 10
x = --------------- = 3 answer//
...............2


........- 4 - 10
x = -------------- = - 7 answer//
.............2
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Vaman say: x^2 + 4x - 21 = 0,
x^2 + 4x +4 = 25 Add and subtract 4 and transfer it.
(x+2)^2=-25
x+2 = +/- i 5. Answer will be
x= -2 + i5, x= -2-i5
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David say: If: x^2 +4x -21 = 0
Then when factored: (x +7)(x -3) = 0
Therefore: x = -7 or x = 3
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la console say: x² + 4x - 21 = 0


Polynomial like: ax² + bx + c, where:
a = 1
b = 4
c = - 21

Δ = b² - 4ac (discriminant)

Δ = (4)² - (4 * 1 * - 21)

Δ = 16 - (- 84)

Δ = 16 + 84

Δ = 100 = 10² → when Δ > 0, you can obtain 2 distinct roots, and these 2 roots are:

x₁ = (- b - √Δ) / 2a = (- 4 - 10) / (2 * 1) = - 14/2 = - 7

x₂ = (- b + √Δ) / 2a = (- 4 + 10) / (2 * 1) = 6/2 = 3
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say: The factored equation is of the form (x + A) * (x + B)

Since A * B has to equal 21, the only choices are 1, 21 and 7,3. There is no way to add 1 and 21 in any method that gives you 4 for the intermediate term. But 7 - 3 = 4

(x + 7) * (x - 3)

x = -7 and x = 3
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Captain Matticus, LandPiratesInc say: a = 1 , b = 4 , c = -21

x = (-b +/- sqrt(b^2 - 4ac)) / (2a)
x = (-4 +/- sqrt(16 + 84)) / 2
x = (-4 +/- sqrt(100)) / 2
x = (-4 +/- 10) / 2
x = -2 +/- 5
x = -7 , 3
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