What are the solutions to the equation d^2-22d+21=0 ?

The solutions to the equation d^2-22d+21=0 are d=21,d=1

Find the zeros of d^2-22d+21=0

The zeros of d^2-22d+21=0 are d=21,d=1

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d^{2} - 22 d + 21 = 0

d = 21, d = 1

Solution steps

d^{2} - 22 d + 21 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 1 \cdot 21}{2 \cdot 1}

(- 22)^{2} - 4 \cdot 1 \cdot 21 = 20

d_{1, 2} = \frac{- ( - 22 ) \pm 20}{2 \cdot 1}

Separate the solutions

d_{1} = \frac{- ( - 22 ) + 20}{2 \cdot 1}, d_{2} = \frac{- ( - 22 ) - 20}{2 \cdot 1}

d = \frac{- ( - 22 ) + 20}{2 \cdot 1} : 21

d = \frac{- ( - 22 ) - 20}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

d = 21, d = 1

(\frac{1}{2})^{4}

\frac{1}{\frac{- 1}{2}}

lo g_{3} (\frac{1}{9}) = x

- 8 ∣ 4 - x ∣ < - 24

\frac{5}{4} \cdot 4

What are the solutions to the equation d^2-22d+21=0 ?
The solutions to the equation d^2-22d+21=0 are d=21,d=1
Find the zeros of d^2-22d+21=0
The zeros of d^2-22d+21=0 are d=21,d=1