Sure! Let's solve the quadratic equation [tex]\( 8t^2 + 20t - 32 = 0 \)[/tex] step-by-step.
1. Identify the coefficients:
In the quadratic equation [tex]\( at^2 + bt + c = 0 \)[/tex], we have:
[tex]\[
a = 8, \quad b = 20, \quad c = -32
\][/tex]
2. Write down the quadratic formula:
The general solution for a quadratic equation [tex]\( at^2 + bt + c = 0 \)[/tex] is given by:
[tex]\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
3. Calculate the discriminant:
The discriminant [tex]\( \Delta \)[/tex] is calculated as:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[
\Delta = 20^2 - 4 \cdot 8 \cdot (-32)
\][/tex]
[tex]\[
\Delta = 400 + 1024
\][/tex]
[tex]\[
\Delta = 1424
\][/tex]
4. Compute the square root of the discriminant:
[tex]\[
\sqrt{1424} = \sqrt{16 \cdot 89} = 4\sqrt{89}
\][/tex]
5. Substitute the values into the quadratic formula:
[tex]\[
t = \frac{-20 \pm 4\sqrt{89}}{2 \cdot 8}
\][/tex]
Simplifying the denominator:
[tex]\[
t = \frac{-20 \pm 4\sqrt{89}}{16}
\][/tex]
Dividing both terms in the numerator by 16:
[tex]\[
t = \frac{-20}{16} \pm \frac{4\sqrt{89}}{16}
\][/tex]
6. Simplify the fractions:
[tex]\[
t = -\frac{20}{16} \pm \frac{4\sqrt{89}}{16}
\][/tex]
[tex]\[
t = -\frac{5}{4} \pm \frac{\sqrt{89}}{4}
\][/tex]
7. Separate the solutions:
[tex]\[
t_1 = -\frac{5}{4} + \frac{\sqrt{89}}{4}
\][/tex]
[tex]\[
t_2 = -\frac{5}{4} - \frac{\sqrt{89}}{4}
\][/tex]
So, the solutions to the quadratic equation [tex]\( 8t^2 + 20t - 32 = 0 \)[/tex] are:
[tex]\[
t = -\frac{5}{4} + \frac{\sqrt{89}}{4} \quad \text{and} \quad t = -\frac{5}{4} - \frac{\sqrt{89}}{4}
\][/tex]
This can also be written as:
[tex]\[
t = -\frac{5}{4} + \frac{\sqrt{89}}{4}, \quad t = -\frac{\sqrt{89}}{4} - \frac{5}{4}
\][/tex]
These represent the two values of [tex]\( t \)[/tex] that satisfy the given quadratic equation.
