Solve 16y^2-8y-1 - ScanMath

Question

Find the roots

y_{1} = \frac{1 - 2}{4}, y_{2} = \frac{1 + 2}{4}

Alternative Form

y_{1} \approx - 0.103553, y_{2} \approx 0.603553

Evaluate

16 y^{2} - 8 y - 1

To find the roots of the expression,set the expression equal to 0

16 y^{2} - 8 y - 1 = 0

Substitute a= 16,b= - 8 and c= - 1 into the quadratic formula y = \frac{- b \pm b ^{2} - 4 a c}{2 a}

y = \frac{8 \pm ( - 8 ) ^{2} - 4 \times 16 ( - 1 )}{2 \times 16}

Simplify the expression

y = \frac{8 \pm ( - 8 ) ^{2} - 4 \times 16 ( - 1 )}{32}

Simplify the expression

More Steps

y = \frac{8 \pm 128}{32}

Simplify the radical expression

More Steps

y = \frac{8 \pm 8 2}{32}

Separate the equation into 2 possible cases

y = \frac{8 + 8 2}{32} y = \frac{8 - 8 2}{32}

Simplify the expression

More Steps

y = \frac{1 + 2}{4} y = \frac{8 - 8 2}{32}

Simplify the expression

More Steps

y = \frac{1 + 2}{4} y = \frac{1 - 2}{4}

Solution

y_{1} = \frac{1 - 2}{4}, y_{2} = \frac{1 + 2}{4}

Alternative Form

y_{1} \approx - 0.103553, y_{2} \approx 0.603553

Show Solution