What are the solutions to the equation 82=(A+B)x^2+(C-B)x+(81A-C) ?

The solutions to the equation 82=(A+B)x^2+(C-B)x+(81A-C) are

Find the zeros of 82=(A+B)x^2+(C-B)x+(81A-C)

The zeros of 82=(A+B)x^2+(C-B)x+(81A-C) are

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82 = (A + B) x^{2} + (C - B) x + (81 A - C)

x = \frac{- C + B + B ^{2} - 324 A B + 2 BC + 328 B - 324 A ^{2} + 328 A + C ^{2} + 4 A C}{2 ( A + B )}, x = \frac{- C + B - B ^{2} - 324 A B + 2 BC + 328 B - 324 A ^{2} + 328 A + C ^{2} + 4 A C}{2 ( A + B )}; A + B \neq = 0

Solution steps

82 = (A + B) x^{2} + (C - B) x + (81 A - C)

Expand

(A + B) x^{2} + (C - B) x + (81 A - C) : A x^{2} + B x^{2} + C x - B x + 81 A - C

82 = A x^{2} + B x^{2} + C x - B x + 81 A - C

Switch sides

A x^{2} + B x^{2} + C x - B x + 81 A - C = 82

Move

82

to the left side

A x^{2} + B x^{2} + C x - B x + 81 A - C - 82 = 0

Write in the standard form

a x^{2} + b x + c = 0

(A + B) x^{2} + (C - B) x + 81 A - C - 82 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( C - B ) \pm ( C - B ) ^{2} - 4 ( A + B ) ( 81 A - C - 82 )}{2 ( A + B )}

Simplify

(C - B)^{2} - 4 (A + B) (81 A - C - 82) : B^{2} - 324 A B + 2 BC + 328 B - 324 A^{2} + 328 A + C^{2} + 4 A C

x_{1, 2} = \frac{- ( C - B ) \pm B ^{2} - 324 A B + 2 BC + 328 B - 324 A ^{2} + 328 A + C ^{2} + 4 A C}{2 ( A + B )}; A + B \neq = 0

Separate the solutions

x_{1} = \frac{- ( C - B ) + B ^{2} - 324 A B + 2 BC + 328 B - 324 A ^{2} + 328 A + C ^{2} + 4 A C}{2 ( A + B )}, x_{2} = \frac{- ( C - B ) - B ^{2} - 324 A B + 2 BC + 328 B - 324 A ^{2} + 328 A + C ^{2} + 4 A C}{2 ( A + B )}

x = \frac{- ( C - B ) + B ^{2} - 324 A B + 2 BC + 328 B - 324 A ^{2} + 328 A + C ^{2} + 4 A C}{2 ( A + B )} : \frac{- C + B + B ^{2} - 324 A B + 2 BC + 328 B - 324 A ^{2} + 328 A + C ^{2} + 4 A C}{2 ( A + B )}

x = \frac{- ( C - B ) - B ^{2} - 324 A B + 2 BC + 328 B - 324 A ^{2} + 328 A + C ^{2} + 4 A C}{2 ( A + B )} : \frac{- C + B - B ^{2} - 324 A B + 2 BC + 328 B - 324 A ^{2} + 328 A + C ^{2} + 4 A C}{2 ( A + B )}

The solutions to the quadratic equation are:

x = \frac{- C + B + B ^{2} - 324 A B + 2 BC + 328 B - 324 A ^{2} + 328 A + C ^{2} + 4 A C}{2 ( A + B )}, x = \frac{- C + B - B ^{2} - 324 A B + 2 BC + 328 B - 324 A ^{2} + 328 A + C ^{2} + 4 A C}{2 ( A + B )}; A + B \neq = 0

5 x^{2} - 34 x = 0

9 x^{2} - 6 x + 8 = 0

0.1 x^{2} - 0.7 x + 1 = 0

4 x^{2} - x - 8 = 0

x^{2} - 2 x + c = 0

What are the solutions to the equation 82=(A+B)x^2+(C-B)x+(81A-C) ?
The solutions to the equation 82=(A+B)x^2+(C-B)x+(81A-C) are
Find the zeros of 82=(A+B)x^2+(C-B)x+(81A-C)
The zeros of 82=(A+B)x^2+(C-B)x+(81A-C) are