(4r+7)(4r-7)=0

Solution for (4r+7)(4r-7)=0 equation:

(4r+7)(4r-7)=0

We use the square of the difference formula

16r^2-49=0

a = 16; b = 0; c = -49;
Δ = b2-4ac
Δ = 02-4·16·(-49)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-56}{2*16}=\frac{-56}{32}
=-1+3/4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+56}{2*16}=\frac{56}{32}
=1+3/4
$