(4t+5)(2t-5)=0

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Solution for (4t+5)(2t-5)=0 equation:



(4t+5)(2t-5)=0
We multiply parentheses ..
(+8t^2-20t+10t-25)=0
We get rid of parentheses
8t^2-20t+10t-25=0
We add all the numbers together, and all the variables
8t^2-10t-25=0
a = 8; b = -10; c = -25;
Δ = b2-4ac
Δ = -102-4·8·(-25)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-30}{2*8}=\frac{-20}{16} =-1+1/4 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+30}{2*8}=\frac{40}{16} =2+1/2 $

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