What are the solutions to the equation (50t)^2+(60t)^2=300^2 ?

The solutions to the equation (50t)^2+(60t)^2=300^2 are t=(30)/(sqrt(61)),t=-(30)/(sqrt(61))

Find the zeros of (50t)^2+(60t)^2=300^2

The zeros of (50t)^2+(60t)^2=300^2 are t=(30)/(sqrt(61)),t=-(30)/(sqrt(61))

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(50 t)^{2} + (60 t)^{2} = 30 0^{2}

t = \frac{30}{61}, t = - \frac{30}{61}

Decimal Notation

t = 3.84110 \dots, t = - 3.84110 \dots

Solution steps

(50 t)^{2} + (60 t)^{2} = 30 0^{2}

Expand

(50 t)^{2} + (60 t)^{2} : 6100 t^{2}

Expand

30 0^{2} : 90000

6100 t^{2} = 90000

Move

90000

to the left side

6100 t^{2} - 90000 = 0

Divide both sides by

6100

\frac{6100 t ^{2}}{6100} - \frac{90000}{6100} = \frac{0}{6100}

Write in the standard form

a x^{2} + b x + c = 0

t^{2} - \frac{900}{61} = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot ( - \frac{900}{61} )}{2 \cdot 1}

0^{2} - 4 \cdot 1 \cdot (- \frac{900}{61}) = \frac{60}{61}

t_{1, 2} = \frac{- 0 \pm \frac{60}{61}}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 0 + \frac{60}{61}}{2 \cdot 1}, t_{2} = \frac{- 0 - \frac{60}{61}}{2 \cdot 1}

t = \frac{- 0 + \frac{60}{61}}{2 \cdot 1} : \frac{30}{61}

t = \frac{- 0 - \frac{60}{61}}{2 \cdot 1} : - \frac{30}{61}

The solutions to the quadratic equation are:

t = \frac{30}{61}, t = - \frac{30}{61}

11 c^{2} + 27 c + 14 = 2 c^{2}

5 x^{2} + 10 x - 2 = 0

x, 36 x^{2} - 36 x + 36 y^{2} + 24 y + 13 = 0

2 - 3 x - 2 x^{2} = 0

m^{2} + 100 m + 2500 = 0

What are the solutions to the equation (50t)^2+(60t)^2=300^2 ?
The solutions to the equation (50t)^2+(60t)^2=300^2 are t=(30)/(sqrt(61)),t=-(30)/(sqrt(61))
Find the zeros of (50t)^2+(60t)^2=300^2
The zeros of (50t)^2+(60t)^2=300^2 are t=(30)/(sqrt(61)),t=-(30)/(sqrt(61))