What are the solutions to the equation 4r^2-8r+12=0 ?

The solutions to the equation 4r^2-8r+12=0 are r=1+sqrt(2)i,r=1-sqrt(2)i

Find the zeros of 4r^2-8r+12=0

The zeros of 4r^2-8r+12=0 are r=1+sqrt(2)i,r=1-sqrt(2)i

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4 r^{2} - 8 r + 12 = 0

r = 1 + 2 i, r = 1 - 2 i

Solution steps

4 r^{2} - 8 r + 12 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 4 \cdot 12}{2 \cdot 4}

Simplify

(- 8)^{2} - 4 \cdot 4 \cdot 12 : 82 i

r_{1, 2} = \frac{- ( - 8 ) \pm 8 2 i}{2 \cdot 4}

Separate the solutions

r_{1} = \frac{- ( - 8 ) + 8 2 i}{2 \cdot 4}, r_{2} = \frac{- ( - 8 ) - 8 2 i}{2 \cdot 4}

r = \frac{- ( - 8 ) + 8 2 i}{2 \cdot 4} : 1 + 2 i

r = \frac{- ( - 8 ) - 8 2 i}{2 \cdot 4} : 1 - 2 i

The solutions to the quadratic equation are:

r = 1 + 2 i, r = 1 - 2 i

3 x + 9 + 4 x + x = 8 x + 5

\frac{x}{4} + 12 = 33

8 x^{2} + 34 x - 30

5 x = 625

∣ x + 2 ∣ > 2

What are the solutions to the equation 4r^2-8r+12=0 ?
The solutions to the equation 4r^2-8r+12=0 are r=1+sqrt(2)i,r=1-sqrt(2)i
Find the zeros of 4r^2-8r+12=0
The zeros of 4r^2-8r+12=0 are r=1+sqrt(2)i,r=1-sqrt(2)i