What are the solutions to the equation 0=4t^2+9+(2t-16)^2 ?

The solutions to the equation 0=4t^2+9+(2t-16)^2 are t=4+i(sqrt(274))/4 ,t=4-i(sqrt(274))/4

Find the zeros of 0=4t^2+9+(2t-16)^2

The zeros of 0=4t^2+9+(2t-16)^2 are t=4+i(sqrt(274))/4 ,t=4-i(sqrt(274))/4

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0 = 4 t^{2} + 9 + (2 t - 16)^{2}

t = 4 + i \frac{274}{4}, t = 4 - i \frac{274}{4}

Solution steps

0 = 4 t^{2} + 9 + (2 t - 16)^{2}

Expand

4 t^{2} + 9 + (2 t - 16)^{2} : 8 t^{2} - 64 t + 265

0 = 8 t^{2} - 64 t + 265

Switch sides

8 t^{2} - 64 t + 265 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 64 ) \pm ( - 64 ) ^{2} - 4 \cdot 8 \cdot 265}{2 \cdot 8}

Simplify

(- 64)^{2} - 4 \cdot 8 \cdot 265 : 4274 i

t_{1, 2} = \frac{- ( - 64 ) \pm 4 274 i}{2 \cdot 8}

Separate the solutions

t_{1} = \frac{- ( - 64 ) + 4 274 i}{2 \cdot 8}, t_{2} = \frac{- ( - 64 ) - 4 274 i}{2 \cdot 8}

t = \frac{- ( - 64 ) + 4 274 i}{2 \cdot 8} : 4 + i \frac{274}{4}

t = \frac{- ( - 64 ) - 4 274 i}{2 \cdot 8} : 4 - i \frac{274}{4}

The solutions to the quadratic equation are:

t = 4 + i \frac{274}{4}, t = 4 - i \frac{274}{4}

∣ u - 4 ∣ < 3

4 x^{2} = 24 x

\frac{3 x - 1}{8} - \frac{2 x + 7}{6} = \frac{5 x}{9}

\frac{16 u ^{3} v}{u v ^{5}}

What are the solutions to the equation 0=4t^2+9+(2t-16)^2 ?
The solutions to the equation 0=4t^2+9+(2t-16)^2 are t=4+i(sqrt(274))/4 ,t=4-i(sqrt(274))/4
Find the zeros of 0=4t^2+9+(2t-16)^2
The zeros of 0=4t^2+9+(2t-16)^2 are t=4+i(sqrt(274))/4 ,t=4-i(sqrt(274))/4