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(4x+3)(2x+16)=35
We move all terms to the left:
(4x+3)(2x+16)-(35)=0
We multiply parentheses ..
(+8x^2+64x+6x+48)-35=0
We get rid of parentheses
8x^2+64x+6x+48-35=0
We add all the numbers together, and all the variables
8x^2+70x+13=0
a = 8; b = 70; c = +13;
Δ = b2-4ac
Δ = 702-4·8·13
Δ = 4484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4484}=\sqrt{4*1121}=\sqrt{4}*\sqrt{1121}=2\sqrt{1121}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-2\sqrt{1121}}{2*8}=\frac{-70-2\sqrt{1121}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+2\sqrt{1121}}{2*8}=\frac{-70+2\sqrt{1121}}{16} $
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