(2t-3)(4t=4)=(2t+7)(4t-3)

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Solution for (2t-3)(4t=4)=(2t+7)(4t-3) equation:



(2t-3)(4t=4)=(2t+7)(4t-3)
We move all terms to the left:
(2t-3)(4t-(4))=0
We multiply parentheses ..
(+8t^2-8t-12t+12)=0
We get rid of parentheses
8t^2-8t-12t+12=0
We add all the numbers together, and all the variables
8t^2-20t+12=0
a = 8; b = -20; c = +12;
Δ = b2-4ac
Δ = -202-4·8·12
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4}{2*8}=\frac{16}{16} =1 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4}{2*8}=\frac{24}{16} =1+1/2 $

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