What are the solutions to the equation 21T^2-320T+448=0 ?

The solutions to the equation 21T^2-320T+448=0 are T=(8(20+sqrt(253)))/(21),T=(8(20-sqrt(253)))/(21)

Find the zeros of 21T^2-320T+448=0

The zeros of 21T^2-320T+448=0 are T=(8(20+sqrt(253)))/(21),T=(8(20-sqrt(253)))/(21)

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21 T^{2} - 320 T + 448 = 0

T = \frac{8 ( 20 + 253 )}{21}, T = \frac{8 ( 20 - 253 )}{21}

Decimal Notation

T = 13.67846 \dots, T = 1.55962 \dots

Solution steps

21 T^{2} - 320 T + 448 = 0

Solve with the quadratic formula

T_{1, 2} = \frac{- ( - 320 ) \pm ( - 320 ) ^{2} - 4 \cdot 21 \cdot 448}{2 \cdot 21}

(- 320)^{2} - 4 \cdot 21 \cdot 448 = 16253

T_{1, 2} = \frac{- ( - 320 ) \pm 16 253}{2 \cdot 21}

Separate the solutions

T_{1} = \frac{- ( - 320 ) + 16 253}{2 \cdot 21}, T_{2} = \frac{- ( - 320 ) - 16 253}{2 \cdot 21}

T = \frac{- ( - 320 ) + 16 253}{2 \cdot 21} : \frac{8 ( 20 + 253 )}{21}

T = \frac{- ( - 320 ) - 16 253}{2 \cdot 21} : \frac{8 ( 20 - 253 )}{21}

The solutions to the quadratic equation are:

T = \frac{8 ( 20 + 253 )}{21}, T = \frac{8 ( 20 - 253 )}{21}

\frac{2 n}{3} + \frac{1}{2} = \frac{2 n - 3}{6}

∣ x - 3 ∣ = ∣ 3 x + 8 ∣

4 x - 5 = 16 + 11 x

x^{2} < 6 x - 8

(y^{\frac{3}{4}})^{\frac{7}{8}}

What are the solutions to the equation 21T^2-320T+448=0 ?
The solutions to the equation 21T^2-320T+448=0 are T=(8(20+sqrt(253)))/(21),T=(8(20-sqrt(253)))/(21)
Find the zeros of 21T^2-320T+448=0
The zeros of 21T^2-320T+448=0 are T=(8(20+sqrt(253)))/(21),T=(8(20-sqrt(253)))/(21)