How do you solve $20 t^{2} - 17 t = 63$ $20t ^ { 2} - 17t = 63$ ?

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How do you solve $20 t^{2} - 17 t = 63$ $20t ^ { 2} - 17t = 63$ ?

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Answer 1 · 2016-12-09T17:06:15

t= $- \frac{7}{5} or \frac{9}{4}$ $-7/5 or 9/4$

Explanation:

This can be solved by either using quadratic formula or by factorising the quadratic $20 t^{2} - 17 t - 63 = 0$ $20t^2 -17t-63=0$ by splitting the middle term. The latter would be quite simpler. Consider the product of the coefficient of $t^{2}$ $t^2$ and the constant term that is 20(-63). Now split this in two parts so that their sum or difference is -17. Now, factorise 20(-63)= 45(-9)7= 28(-45). The sum of 28and -45 gives -17. Thus write the quadratic as $20 t^{2} - 45 t + 28 t - 63 = 0$ $20t^2 -45t +28t -63=0$ . Now pair the terms as follows

#(20t^2 -45t) +(28t-63)=0

$5 t (4 t - 9) + 7 (4 t - 9) = 0$ $5t(4t-9) +7(4t-9)=0$
(5t+7)(4t-9)=0
7
This gives t= $- \frac{7}{5}$ $-7/5$ or t= $\frac{9}{4}$ $9/4$

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How do you solve $20 t^{2} - 17 t = 63$ $20t ^ { 2} - 17t = 63$ ?

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How do you solve 20t2−17t=6320t ^ { 2} - 17t = 63?

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How do you solve $20 t^{2} - 17 t = 63$ $20t ^ { 2} - 17t = 63$ ?