What are the solutions to the equation a^2-10a+25=49 ?

The solutions to the equation a^2-10a+25=49 are a=12,a=-2

Find the zeros of a^2-10a+25=49

The zeros of a^2-10a+25=49 are a=12,a=-2

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a^{2} - 10 a + 25 = 49

a = 12, a = - 2

Solution steps

a^{2} - 10 a + 25 = 49

Move

49

to the left side

a^{2} - 10 a - 24 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot ( - 24 )}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot (- 24) = 14

a_{1, 2} = \frac{- ( - 10 ) \pm 14}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- ( - 10 ) + 14}{2 \cdot 1}, a_{2} = \frac{- ( - 10 ) - 14}{2 \cdot 1}

a = \frac{- ( - 10 ) + 14}{2 \cdot 1} : 12

a = \frac{- ( - 10 ) - 14}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

a = 12, a = - 2

\frac{x ^{- 1} + 1}{y x ^{- 1} + x}

\frac{3 x ^{3}}{6 y ^{2}} \cdot \frac{4 y ^{5}}{7 x ^{4}}

∣ y ∣ - 3 \leq 0

(y^{2})^{4}

\frac{x - 3}{x + 1} < 1

What are the solutions to the equation a^2-10a+25=49 ?
The solutions to the equation a^2-10a+25=49 are a=12,a=-2
Find the zeros of a^2-10a+25=49
The zeros of a^2-10a+25=49 are a=12,a=-2