What is the answer to solvefor a,b^2-4ac=0 ?

The answer to solvefor a,b^2-4ac=0 is a=(b^2)/(4c);\quad c\ne 0

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solvefor a,b^2-4ac=0

solve for

a, b^{2} - 4 a c = 0

a = \frac{b ^{2}}{4 c}; c \neq = 0

Solution steps

b^{2} - 4 a c = 0

Move

b^{2}

to the right side

- 4 a c = - b^{2}

Divide both sides by

- 4 c; c \neq = 0

a = \frac{b ^{2}}{4 c}; c \neq = 0

13 + \frac{x}{4} = x + 1

y, 5 x + y = 7

y, 4 x + 2 y = - 2

y, 4 x + 2 y = 12

h, t = 2 h + 2 u

What is the answer to solvefor a,b^2-4ac=0 ?
The answer to solvefor a,b^2-4ac=0 is a=(b^2)/(4c);\quad c\ne 0