t^2-15t+48=0

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Solution for t^2-15t+48=0 equation: 



t^2-15t+48=0

a = 1; b = -15; c = +48;
Δ = b2-4ac
Δ = -152-4·1·48
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{33}}{2*1}=\frac{15-\sqrt{33}}{2}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{33}}{2*1}=\frac{15+\sqrt{33}}{2}
$

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