What are the solutions to the equation 16x^2-48x+36=49 ?

The solutions to the equation 16x^2-48x+36=49 are x= 13/4 ,x=-1/4

Find the zeros of 16x^2-48x+36=49

The zeros of 16x^2-48x+36=49 are x= 13/4 ,x=-1/4

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16 x^{2} - 48 x + 36 = 49

x = \frac{13}{4}, x = - \frac{1}{4}

Decimal Notation

x = 3.25, x = - 0.25

Solution steps

16 x^{2} - 48 x + 36 = 49

Move

49

to the left side

16 x^{2} - 48 x - 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 \cdot 16 ( - 13 )}{2 \cdot 16}

(- 48)^{2} - 4 \cdot 16 (- 13) = 56

x_{1, 2} = \frac{- ( - 48 ) \pm 56}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 48 ) + 56}{2 \cdot 16}, x_{2} = \frac{- ( - 48 ) - 56}{2 \cdot 16}

x = \frac{- ( - 48 ) + 56}{2 \cdot 16} : \frac{13}{4}

x = \frac{- ( - 48 ) - 56}{2 \cdot 16} : - \frac{1}{4}

The solutions to the quadratic equation are:

x = \frac{13}{4}, x = - \frac{1}{4}

(x + 2) (x - 3) < 0

20 x^{2} - 980 x

(\frac{5}{7})^{3}

\frac{3 ^{3}}{5}

w^{2} - 9 = 0

What are the solutions to the equation 16x^2-48x+36=49 ?
The solutions to the equation 16x^2-48x+36=49 are x= 13/4 ,x=-1/4
Find the zeros of 16x^2-48x+36=49
The zeros of 16x^2-48x+36=49 are x= 13/4 ,x=-1/4