What are the solutions to the equation 4b^2=4b+48 ?

The solutions to the equation 4b^2=4b+48 are b=4,b=-3

Find the zeros of 4b^2=4b+48

The zeros of 4b^2=4b+48 are b=4,b=-3

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4 b^{2} = 4 b + 48

b = 4, b = - 3

Solution steps

4 b^{2} = 4 b + 48

Move

48

to the left side

4 b^{2} - 48 = 4 b

Move

4 b

to the left side

4 b^{2} - 48 - 4 b = 0

Write in the standard form

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

4 b^{2} - 4 b - 48 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 4 ( - 48 )}{2 \cdot 4}

(- 4)^{2} - 4 \cdot 4 (- 48) = 28

b_{1, 2} = \frac{- ( - 4 ) \pm 28}{2 \cdot 4}

Separate the solutions

b_{1} = \frac{- ( - 4 ) + 28}{2 \cdot 4}, b_{2} = \frac{- ( - 4 ) - 28}{2 \cdot 4}

b = \frac{- ( - 4 ) + 28}{2 \cdot 4} : 4

b = \frac{- ( - 4 ) - 28}{2 \cdot 4} : - 3

The solutions to the quadratic equation are:

b = 4, b = - 3

2 m^{2} - 15 m + 10 = 3 m - m^{2} - 5

4 x^{2} + 18 x - 30 = 0

4 x = - 4 x^{2}

x^{2} - 2 x - 1 = 0

x, f (2 x + 1) = 1 - (x - 3)^{2}

What are the solutions to the equation 4b^2=4b+48 ?
The solutions to the equation 4b^2=4b+48 are b=4,b=-3
Find the zeros of 4b^2=4b+48
The zeros of 4b^2=4b+48 are b=4,b=-3