4r^2+3r-7=0

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Solution for 4r^2+3r-7=0 equation: 



4r^2+3r-7=0

a = 4; b = 3; c = -7;
Δ = b2-4ac
Δ = 32-4·4·(-7)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*4}=\frac{-14}{8}
=-1+3/4
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*4}=\frac{8}{8}
=1
$

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