Method 1:
Completing the Square
x2 - 12x + 16 = 0
x2 - 12x + 16 + 20 = 0 + 20
x2 - 12x + 36 = 20
(x - 6)2 = 20
√((x - 6)2) = ±√(20)
√((x - 6)2) = ±√(2*2*5)
x - 6 = ±2√(5)
x - 6 + 6 = 6 ±2√(5)
x - 6 + 6 = 6 ±2√(5)
x = 6 ± 2√(5)
Method 2:
Quadratic Formula
Standard form of a quadratic equation:
ax2 + bx + c = 0
x2 - 12x + 16 = 0
a = 1
b = -12
c = 16
x = (-b ± √(b2 - 4ac))/ 2a
x = (-(-12) ± √((-12)2 - 4(1)(16)))/ 2(1)
x = ((12) ± √(144 - 64))/ 2
x = (12 ± √(80))/ 2
x = (12 ± 4√(5))/ 2
x = 12/2 ± (4/2)√(5)
x = 6 ± 2√(5)
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If you mean t(t+7) = 4(3+2t), then it is factored. If you meant solve for 't', then the solution is t = t2 - 12.
This is a hyperbola. It is best approached using Fermat's factorisation method. Seefermat-s-factorization-methodor google wikepedia. I don't know of any faster approach.
(p + 12)(p - 7) p = 7, -12
Assuming the question is written as: x2+11x-12 This would factor to: (x+12)(x-1)
4 and 3 are factors of 12.